Class 11

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Miscellaneous Exercise 6 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Miscellaneous Exercise 6

(I) Choose the correct alternative.

Question 1.
Equation of a circle which passes through (3, 6) and touches the axes is
(A) x² + y² + 6x + 6y + 3 = 0
(B) x² + y² – 6x – 6y – 9 = 0
(C) x² + y² – 6x – 6y + 9 = 0
(D) x² + y² – 6x + 6y – 3 = 0
Answer:
(C) x² + y² – 6x – 6y + 9 = 0

Question 2.
If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 sq. units, then find the equation of the circle.
(A) x² + y² – 2x + 2y = 40
(B) x² + y² – 2x – 2y = 47
(C) x² + y² – 2x + 2y = 47
(D) x² + y² – 2x – 2y = 40
Answer:
(C) x² + y² – 2x + 2y = 47
Hint:
Centre of circle = Point of intersection of diameters.
Solving 2x – 3y = 5 and 3x – 4y = 7, we get
x = 1, y = -1
Centre of the circle C(h, k) = C(1, -1)
∴ Area = 154
πr² = 154
\(\frac{22}{7} \times r^{2}\) = 154
r² = 154 × \(\frac{22}{7}\) = 49
∴ r = 7
equation of the circle is
(x – 1)² + (y + 1)² = 72
x² + y² – 2x + 2y = 47

Question 3.
Find the equation of the circle which passes through the points (2, 3) and (4, 5), and the center lies on the straight line y – 4x + 3 = 0.
(A) x² + y² – 4x – 10y + 25 = 0
(B) x² + y² – 4x – 10y – 25 = 0
(C) x² + y² – 4x + 10y – 25 = 0
(D) x² + y² + 4x – 10y + 25 = 0
Answer:
(A) x² + y² – 4x – 10y + 25 = 0

Question 4.
The equation(s) of the tangent(s) to the circle x² + y² = 4 which are parallel to x + 2y + 3 = 0 are
(A) x – 2y = 2
(B) x + 2y = ±2√3
(C) x + 2y = ±2√5
(D) x – 2y = ±2√5
Answer:
(C) x + 2y = ±2√5

Question 5.
If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find the radius of the circle.
(A) \(\frac{3}{4}\)
(B) \(\frac{4}{3}\)
(C) \(\frac{1}{4}\)
(D) \(\frac{7}{4}\)
Answer:
(A) \(\frac{3}{4}\)
Hint:
Tangents are parallel to each other.
The perpendicular distance between tangents = diameter

Question 6.
The area of the circle having centre at (1, 2) and passing through (4, 6) is
(A) 5π
(B) 10π
(C) 25π
(D) 100π
Answer:
(C) 25π
Hint:

Question 7.
If a circle passes through the points (0, 0), (a, 0), and (0, b), then find the co-ordinates of its centre.
(A) \(\left(\frac{-a}{2}, \frac{-b}{2}\right)\)
(B) \(\left(\frac{a}{2}, \frac{-b}{2}\right)\)
(C) \(\left(\frac{-a}{2}, \frac{b}{2}\right)\)
(D) \(\left(\frac{a}{2}, \frac{b}{2}\right)\)
Answer:
(D) \(\left(\frac{a}{2}, \frac{b}{2}\right)\)

Question 8.
The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is
(A) x² + y² = 9a²
(B) x² + y² = 16a²
(C) x² + y² = 4a²
(D) x² + y² = a²
Answer:
(C) x² + y² = 4a²
Hint:
Since the triangle is equilateral.
The centroid of the triangle is same as the circumcentre
and radius of the circumcircle = \(\frac{2}{3}\) (median) = \(\frac{2}{3}\)(3a) = 2a
Hence, the equation of the circumcircle whose centre is at (0, 0) and radius 2a is x² + y² = 4a²

Question 9.
A pair of tangents are drawn to a unit circle with centre at the origin and these tangents intersect at A enclosing an angle of 60. The area enclosed by these tangents and the arc of the circle is
(A) \(\frac{2}{\sqrt{3}}-\frac{\pi}{6}\)
(B) \(\sqrt{3}-\frac{\pi}{3}\)
(C) \(\frac{\pi}{3}-\frac{\sqrt{3}}{6}\)
(D) \(\sqrt{3}\left(1-\frac{\pi}{6}\right)\)
Answer:
(B) \(\sqrt{3}-\frac{\pi}{3}\)
Hint:

Question 10.
The parametric equations of the circle x² + y² + mx + my = 0 are
(A) x = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \cos \theta\), y = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \sin \theta\)
(B) x = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \cos \theta\), y = \(\frac{+m}{2}+\frac{m}{\sqrt{2}} \sin \theta\)
(C) x = 0, y = 0
(D) x = m cos θ, y = m sin θ
Answer:
(A) x = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \cos \theta\), y = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \sin \theta\)

(II) Answer the following:

Question 1.
Find the centre and radius of the circle x² + y² – x + 2y – 3 = 0.
Solution:
Given equation of the circle is x² + y² – x + 2y – 3 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -1, 2f = 2 and c = -3
g = \(\frac{-1}{2}\), f = 1 and c = -3
Centre of the circle = (-g, -f) = (\(\frac{1}{2}\), -1)
and radius of the circle

Question 2.
Find the centre and radius of the circle x = 3 – 4 sin θ, y = 2 – 4 cos θ.
Solution:
Given, x = 3 – 4 sin θ, y = 2 – 4 cos θ
⇒ x – 3 = -4 sin θ, y – 2 = -4 cos θ
On squaring and adding, we get
⇒ (x – 3)² + (y – 2)² = (-4 sin θ)² + (-4 cos θ)²
⇒ (x – 3)² + (y – 2)² = 16 sin² θ + 16 cos² θ
⇒ (x – 3)² + (y – 2)² = 16(sin² θ + cos² θ)
⇒ (x – 3)² + (y – 2)² = 16(1)
⇒ (x – 3)² + (y – 2)² = 16
⇒ (x – 3)² + (y – 2)² = 4²
Comparing this equation with (x – h)² + (y – k)² = r², we get
h = 3, k = 2, r = 4
∴ Centre of the circle is (3, 2) and radius is 4.

Question 3.
Find the equation of circle passing through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0 and whose centre is the point of intersection of lines x + y + 1 = 0 and x – 2y + 4 = 0.
Solution:
Required circle passes through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0.
x + 3y = 0
⇒ x = -3y ……..(i)
2x – 7y = 0 ……(ii)
Substituting x = -3y in (ii), we get
⇒ 2(-3y) – 7y = 0
⇒ -6y – 7y = 0
⇒ -13y = 0
⇒ y = 0
Substituting y = 0 in (i), we get
x = -3(0) = 0
Point of intersection is O(0, 0).
This point O(0, 0) lies on the circle.
Let C(h, k) be the centre of the required circle.
Since, point of intersection of lines x + y = -1 and x – 2y = -4 is the centre of circle.
∴ x = h, y = k
∴ Equations of lines become
h + k = -1 ……(iii)
h – 2k = -4 …..(iv)
By (iii) – (iv), we get
3k = 3
⇒ k = 1
Substituting k = 1 in (iii), we get
h + 1 = -1
⇒ h = -2
∴ Centre of the circle is C(-2, 1) and it passes through point O(0, 0).
Radius(r) = OC
= \(\sqrt{(0+2)^{2}+(0-1)^{2}}\)
= \(\sqrt{4+1}\)
= √5
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = -2, k = 1
the required equation of the circle is
(x + 2)² + (y – 1)² = (√5)²
⇒ x² + 4x + 4 + y² – 2y + 1 = 5
⇒ x² + y² + 4x – 2y = 0

Question 4.
Find the equation of the circle which passes through the origin and cuts off chords of lengths 4 and 6 on the positive side of the X-axis and Y-axis respectively.
Solution:

Let the circle cut the chord of length 4 on X-axis at point A and the chord of length 6 on the Y-axis at point B.
∴ the co-ordinates of point A are (4, 0) and co-ordinates of point B are (0, 6).
Since ∠BOA is a right angle.
AB represents the diameter of the circle.
The equation of a circle having (x₁, y₁) and (x₂, y₂) as endpoints of diameter is given by
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
Here, x₁ = 4, y₁ = 0, x₂ = 0, y₂ = 6
∴ the required equation of the circle is
⇒ (x – 4) (x – 0) + (y – 0) (y – 6) = 0
⇒ x² – 4x + y² – 6y = 0
⇒ x² + y² – 4x – 6y = 0

Question 5.
Show that the points (9, 1), (7, 9), (-2, 12) and (6, 10) are concyclic.
Solution:
Let the equation of circle passing through the points (9, 1), (7, 9), (-2, 12) be
x² + y² + 2gx + 2fy + c = 0 …….(i)
For point (9, 1),
Substituting x = 9 andy = 1 in (i), we get
81 + 1 + 18g + 2f + c = 0
⇒ 18g + 2f + c = -82 …..(ii)
For point (7, 9),
Substituting x = 7 andy = 9 in (i), we get
49 + 81 + 14g + 18f + c = 0
⇒ 14g + 18f + c = -130 ……(iii)
For point (-2, 12),
Substituting x = -2 and y = 12 in (i), we get
4 + 144 – 4g + 24f + c = 0
⇒ -4g + 24f + c = -148 …..(iv)
By (ii) – (iii), we get
4g – 16f = 48
⇒ g – 4f = 12 …..(v)
By (iii) – (iv), we get
18g – 6f = 18
⇒ 3g – f = 3 ……(vi)
By 3 × (v) – (vi), we get
-11f = 33
⇒ f = -3
Substituting f = -3 in (vi), we get
3g – (-3) = 3
⇒ 3g + 3 = 3
⇒ g = 0
Substituting g = 0 and f = -3 in (ii), we get
18(0) + 2(-3) + c = – 82
⇒ -6 + c = -82
⇒ c = -76
Equation of the circle becomes
x² + y² + 2(0)x + 2(-3)y + (-76) = 0
⇒ x² + y² – 6y – 76 = 0 ……(vii)
Now for the point (6, 10),
Substituting x = 6 and y = 10 in L.H.S. of (vii), we get
L.H.S = 6² + 10² – 6(10) – 76
= 36 + 100 – 60 – 76
= 0
= R.H.S.
∴ Point (6,10) satisfies equation (vii).
∴ the given points are concyciic.

Question 6.
The line 2x – y + 6 = 0 meets the circle x² + y² + 10x + 9 = 0 at A and B. Find the equation of circle with AB as diameter. Solution:
2x – y + 6 = 0
⇒ y = 2x + 6
Substituting y = 2x + 6 in x² + y² + 10x + 9 = 0, we get
⇒ x² + (2x + 6)² + 10x + 9 = 0
⇒ x² + 4x² + 24x + 36 + 10x + 9 = 0
⇒ 5x² + 34x + 45 = 0
⇒ 5x² + 25x + 9x + 45 = 0
⇒ (5x + 9) (x + 5) = 0
⇒ 5x = -9 or x = -5
⇒ x = \(\frac{-9}{5}\) or x = -5
When x = \(\frac{-9}{5}\),
y = 2 × \(\frac{-9}{5}\) + 6
= \(\frac{-18}{5}\) + 6
= \(\frac{-18+30}{5}\)
= \(\frac{12}{5}\)
∴ Point of intersection is A\(\left(\frac{-9}{5}, \frac{12}{5}\right)\)
When x = -5,
y = -10 + 6 = -4
∴ Point of intersection in B (-5, -4).
By diameter form, equation of circle with AB as diameter is
(x + \(\frac{9}{5}\)) (x + 5) + (y – \(\frac{12}{5}\)) (y + 4) = 0
⇒ (5x + 9) (x + 5) + (5y – 12) (y + 4) = 0
⇒ 5x² + 25x + 9x + 45 + 5y² + 20y – 12y – 48 = 0
⇒ 5x² + 5y² + 34x + 8y – 3 = 0

Question 7.
Show that x = -1 is a tangent to circle x² + y² – 4x – 2y – 4 = 0 at (-1, 1).
Solution:
Given equation of circle is x² + y² – 4x – 2y – 4 = 0.
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -4, 2f = -2, c = -4
⇒ g = -2, f = -1, c = -4
The equation of a tangent to the circle
x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
the equation of the tangent at (-1, 1) is
⇒ x(-1) + y(1) – 2(x – 1) – 1(y + 1) – 4 = 0
⇒ -3x – 3 = 0
⇒ -x – 1 = 0
⇒ x = -1
∴ x = -1 is the tangent to the given circle at (-1, 1).

Question 8.
Find the equation of tangent to the circle x² + y² = 64 at the point P(\(\frac{2 \pi}{3}\)).
Solution:
Given equation of circle is x² + y² = 64
Comparing this equation with x² + y² = r², we get r = 8
The equation of a tangent to the circle x² + y² = r² at P(θ) is x cos θ + y sin θ = r
∴ the equation of the tangent at P(\(\frac{2 \pi}{3}\)) is
⇒ x cos \(\frac{2 \pi}{3}\) + y sin \(\frac{2 \pi}{3}\) = 9
⇒ \(x\left(\frac{-1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=8\)
⇒ -x + √3y = 16
⇒ x – √3y + 16 = 0

Question 9.
Find the equation of locus of the point of intersection of perpendicular tangents drawn to the circle x = 5 cos θ and y = 5 sin θ.
Solution:
The locus of the point of intersection of perpendicular tangents is the director circle of the given circle.
x = 5 cos θ and y = 5 sin θ
⇒ x² + y² = 25 cos² θ + 25 sin² θ
⇒ x² + y² = 25 (cos² θ + sin² θ)
⇒ x² + y² = 25(1) = 25
The equation of the director circle of the circle x² + y² = a² is x² + y² = 2a².
Here, a = 5
∴ the required equation is
x² + y² = 2(5)² = 2(25)
∴ x² + y² = 50

Question 10.
Find the equation of the circle concentric with x² + y² – 4x + 6y = 1 and having radius 4 units.
Solution:
Given equation of circle is
x² + y² – 4x + 6y = 1 i.e., x² + y² – 4x + 6y – 1 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -4, 2f = 6
⇒ g = -2, f = 3
Centre of the circle = (-g, -f) = (2, -3)
Given circle is concentric with the required circle.
∴ They have same centre.
∴ Centre of the required circle = (2, -3)
The equation of a circle with centre at (h, k) and radius r is (x – h)² + (y – k)² = r²
Here, h = 2, k = -3 and r = 4
∴ the required equation of the circle is
(x – 2)² + [y – (-3)]² = 4²
⇒ (x – 2)² + (y + 3)² = 16
⇒ x² – 4x + 4 + y² + 6y + 9 – 16 = 0
⇒ x² + y² – 4x + 6y – 3 = 0

Question 11.
Find the lengths of the intercepts made on the co-ordinate axes, by the circles.
(i) x² + y² – 8x + y – 20 = 0
(ii) x² + y² – 5x + 13y – 14 = 0
Solution:
To find x-intercept made by the circle x² + y² + 2gx + 2fy + c = 0,
substitute y = 0 and get a quadratic equation in x, whose roots are, say, x₁ and x₂.

These values represent the abscissae of ends A and B of the x-intercept.
Length of x-intercept = |AB| = |x₂ – x₁|
Similarly, substituting x = 0, we get a quadratic equation in y whose roots, say, y₁ and y₂ are ordinates of the ends C and D of the y-intercept.
Length of y-intercept = |CD| = |y₂ – y₁|
(i) Given equation of the circle is
x² + y² – 8x + y – 20 = 0 ……(i)
Substituting y = 0 in (i), we get
x² – 8x – 20 = 0 ……(ii)
Let AB represent the x-intercept, where
A = (x₁, 0), B = (x₂, 0)
Then from (ii),
x₁ + x₂ = 8 and x₁x₂ = -20
(x₁ – x₂)² = (x₁ + x₂)² – 4x₁x₂
= (8)² – 4(-20)
= 64 + 80
= 144
∴ |x₁ – x₂| = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}}\) = √144 = 12
∴ Length of x – intercept =12 units
Substituting x = 0 in (i), we get
y² + y – 20 = 0 …..(iii)
Let CD represent the y – intercept,
where C = (0, y₁) and D = (0, y₂)
Then from (iii),
y₁ + y₂ = -1 and y₁y₂ = -20
(y₁ – y₂)² = (y₁ + y₂)² – 4y₁y₂
= (-1)² – 4(-20)
= 1 + 80
= 81
∴ |y₁ – y₂| = \(\sqrt{\left(y_{1}-y_{2}\right)^{2}}\) = √81 = 9
∴ Length of y – intercept = 9 units.

Alternate Method:
Given equation of the circle is x² + y² – 8x + y – 20 = 0 ……(i)
x-intercept:
Substituting y = 0 in (i), we get
x² – 8x – 20 = 0
⇒ (x – 10)(x + 2) = 0
⇒ x = 10 or x = -2
length of x-intercept = |10 – (-2)| = 12 units
y-intercept:
Substituting x = 0 in (i), we get
y² + y – 20 = 0
⇒ (y + 5)(y – 4) = 0
⇒ y = -5 or y = 4
length of y-intercept = |-5 – 4| = 9 units

(ii) Given equation of the circle is
x² + y² – 5x + 13y – 14 = 0
Substituting y = 0 in (i), we get
x² – 5x – 14 = 0 ……(ii)
Let AB represent the x-intercept, where
A = (x₁, 0), B = (x₂, 0)
Then from (ii),
x₁ + x₂ = 5 and x₁x₂ = -14
(x₁ – x₂)² = (x₁ + x₂)² – 4x₁x₂
= (5)² – 4(-14)
= 25 + 56
= 81
∴ |x₁ – x₂| = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}}\) = √81 = 9
∴ Length of x-intercept = 9 units
Substituting x = 0 in (i), we get
y² + 13y – 14 = 0 ……(iii)
Let CD represent they-intercept,
where C = (0, y₁), D = (0, y₂).
Then from (iii),
y₁ + y₂ = -13 and y₁y₂ = -14
(y₁ – y₂)² = (y₁ + y₂)² – 4y₁y₂
= (-13)² – 4(-14)
= 169 + 56
= 225
∴ |y₁ – y₂| = \(\sqrt{\left(y_{1}-y_{2}\right)^{2}}\) = √225 = 15
∴ Length ofy-intercept = 15 units

Question 12.
Show that the circles touch each other externally. Find their point of contact and the equation of their common tangent.
(i) x² + y² – 4x + 10y + 20 = 0
x² + y² + 8x – 6y – 24 = 0
(ii) x² + y² – 4x – 10y + 19 = 0
x² + y² + 2x + 8y – 23 = 0
Solution:
(i) Given equation of the first circle is x² + y² – 4x + 10y + 20 = 0
Here, g = -2, f = 5, c = 20
Centre of the first circle is C₁ = (2, -5)
Radius of the first circle is
r₁ = \(\sqrt{(-2)^{2}+5^{2}-20}\)
= \(\sqrt{4+25-20}\)
= √9
= 3
Given equation of the second circle is x² + y² + 8x – 6y – 24 = 0
Here, g = 4, f = -3, c = -24
Centre of the second circle is C₂ = (-4, 3)
Radius of the second circle is
r₂ = \(\sqrt{4^{2}+(-3)^{2}+24}\)
= \(\sqrt{16+9+24}\)
= √49
= 7
By distance formula,
C₁C₂ = \(\sqrt{(-4-2)^{2}+[3-(-5)]^{2}}\)
= \(\sqrt{36+64}\)
= √1oo
= 10
r₁ + r₂ = 3 + 7 = 10
Since, C₁C₂ = r₁ + r₂
∴ the given circles touch each other externally.

Let P(x, y) be the point of contact.
∴ P divides C₁C₂ internally in the ratio r₁ : r₂ i.e. 3 : 7.
∴ By internal division,

Equation of common tangent is
(x² + y² – 4x + 10y + 20) – (x² + y² + 8x – 6y – 24) = 0
⇒ -4x + 10y + 20 – 8x + 6y + 24 = 0
⇒ -12x + 16y + 44 = 0
⇒ 3x – 4y – 11 = 0

(ii) Given equation of the first circle is x² + y² – 4x – 10y + 19 = 0
Here, g = -2, f = -5, c = 19
Centre of the first circle is C₁ = (2, 5)
Radius of the first circle is
r₁ = \(\sqrt{(-2)^{2}+(-5)^{2}-19}\)
= \(\sqrt{4+25-19}\)
= √10
Given equation of the second circle is x² + y² + 2x + 8y – 23 = 0
Here, g = 1, f = 4, c = -23
Centre of the second circle is C₂ = (-1, -4)
Radius of the second circle is
r₂ = \(\sqrt{(-1)^{2}+4^{2}+23}\)
= \(\sqrt{1+16+23}\)
= √40
= 2√10
By distance formula,
C₁C₂ = \(\sqrt{(-1-2)^{2}+(-4-5)^{2}}\)
= \(\sqrt{9+81}\)
= √90
= 3√10
r₁ + r₂ = √10 + 2√10 = 3√10
Since, C₁C₂ = r₁ + r₂
the given circles touch each other externally.
r₁ : r₂ = √10 : 2√10 = 1 : 2
Let P(x, y) be the point of contact.

∴ P divides C₁ C₂ internally in the ratio r₁ : r₂ i.e. 1 : 2
∴ By internal division,

Point of contact = (1, 2)
Equation of common tangent is
(x² + y² – 4x – 10y + 19) – (x² + y² + 2x + 8y – 23) = 0
⇒ -4x – 10y + 19 – 2x – 8y + 23 = 0
⇒ -6x – 18y + 42 = 0
⇒ x + 3y – 7 = 0

Question 13.
Show that the circles touch each other internally. Find their point of contact and the equation of their common tangent.
(i) x² + y² – 4x – 4y – 28 = 0,
x² + y² – 4x – 12 = 0
(ii) x² + y² + 4x – 12y + 4 = 0,
x² + y² – 2x – 4y + 4 = 0
Solution:
(i) Given equation of the first circle is x² + y² – 4x – 4y – 28 = 0
Here, g = -2, f = -2, c = -28
Centre of the first circle is C₁ = (2, 2)
Radius of the first circle is
r₁ = \(\sqrt{(-2)^{2}+(-2)^{2}+28}\)
= \(\sqrt{4+4+28}\)
= √36
= 6
Given equation of the second circle is x² + y² – 4x – 12 = 0
Here, g = -2, f = 0, c = -12
Centre of the second circle is C₂ = (2, 0)
Radius of the second circle is
r₂ = \(\sqrt{(-2)^{2}+0^{2}+12}\)
= \(\sqrt{4+12}\)
= √16
= 4
By distance formula,
C₁C₂ = \(\sqrt{(2-2)^{2}+(0-2)^{2}}\)
= √4
= 2
|r₁ – r₂| = 6 – 4 = 2
Since, C₁C₂ = |r₁ – r₂|
∴ the given circles touch each other internally.
Equation of common tangent is
(x² + y² – 4x – 4y – 28) – (x² + y² – 4x – 12) = 0
⇒ -4x – 4y – 28 + 4x + 12 = 0
⇒ -4y – 16 = 0
⇒ y + 4 = 0
⇒ y = -4
Substituting y = -4 in x² + y² – 4x – 12 = 0, we get
⇒ x² + (-4)² – 4x – 12 = 0
⇒ x² + 16 – 4x – 12 = 0
⇒ x² – 4x + 4 = 0 .
⇒ (x – 2)² = 0
⇒ x = 2
∴ Point of contact is (2, -4) and equation of common tangent is y + 4 = 0.

(ii) Given equation of the first circle is x² + y² + 4x – 12y + 4 = 0
Here, g = 2, f = -6, c = 4
Centre of the first circle is C₁ = (-2, 6)
Radius of the first circle is
r₁ = \(\sqrt{2^{2}+(-6)^{2}-4}\)
= \(\sqrt{4+36-4}\)
= √36
= 6
Given equation of the second circle is x² + y² – 2x – 4y + 4 = 0
Here, g = -1, f = -2, c = 4
Centre of the second circle is C₂ = (1, 2)
Radius of the second circle is
r₂ = \(\sqrt{(-1)^{2}+(-2)^{2}-4}\)
= \(\sqrt{1+4-4}\)
= √1
= 1
By distance formula,
C₁C₂ = \(\sqrt{[1-(-2)]^{2}+(2-6)^{2}}\)
= \(\sqrt{9+16}\)
= √25
= 5
|r₁ – r₂| = 6 – 1 = 5
Since, C₁C₂ = |r₁ – r₂|
the given circles touch each other internally.
Equation of common tangent is
(x² + y² + 4x – 12y + 4) – (x² + y² – 2x – 4y + 4) = 0
⇒ 4x – 12y + 4 + 2x + 4y – 4 = 0
⇒ 6x – 8y = 0
⇒ 3x – 4y = 0
⇒ y = \(\frac{3 x}{4}\)
Substituting y = \(\frac{3 x}{4}\) in x² + y² – 2x – 4y + 4 = 0, we get

∴ Point of contact is \(\left(\frac{8}{5}, \frac{6}{5}\right)\) and equation of common tangent is 3x – 4y = 0.

Question 14.
Find the length of the tangent segment drawn from the point (5, 3) to the circle x² + y² + 10x – 6y – 17 = 0.
Solution:
Given equation of circle is x² + y² + 10x – 6y – 17 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = 10, 2f = -6, c = -17
⇒ g = 5, f = -3, c = -17
Centre of circle = (-g, -f) = (-5, 3)

In right angled ∆ABC,
BC² = AB² + AC² …..[Pythagoras theorem]
⇒ (10)² = AB²+ (√51)²
⇒ AB² = 100 – 51 = √49
⇒ AB = 7
∴ Length of the tangent segment from (5, 3) is 7 units.

Alternate method:
Given equation of circle is x² + y² + 10x – 6y – 17 = 0
Here, g = 5, f = -3, c = -17
Length of the tangent segment to the circle x² + y² + 2gx + 2fy + c = 0 from the point (x₁, y₁) is \(\sqrt{x_{1}^{2}+y_{1}^{2}+2 g x_{1}+2 f y_{1}+c}\)
Length of the tangent segment from (5, 3)
= \(\sqrt{(5)^{2}+(3)^{2}+10(5)-6(3)-17}\)
= \(\sqrt{25+9+50-18-17}\)
= √49
= 7 units

Question 15.
Find the value of k, if the length of the tangent segment from the point (8, -3) to the circle x² + y² – 2x + ky – 23 = 0 is √10.
Solution:
Given equation of the circle is x² + y² – 2x + ky – 23 = 0
Here, g = -1, f = \(\frac{\mathrm{k}}{2}\), c = -23
Length of the tangent segment to the circle x² + y² + 2gx + 2fy + c = 0 from the point (x₁, y₁) is \(\sqrt{x_{1}^{2}+y_{1}^{2}+2 g x_{1}+2 f y_{1}+c}\)
Length of the tangent segment from (8, -3) = √10
⇒ \(\sqrt{8^{2}+(-3)^{2}-2(8)+k(-3)-23}=\sqrt{10}\)
⇒ 64 + 9 – 16 – 3k – 23 = 10 …..[Squaring both the sides]
⇒ 34 – 3k = 10
⇒ 3k = 24
⇒ k = 8

Question 16.
Find the equation of tangent to circle x² + y² – 6x – 4y = 0, at the point (6, 4) on it.
Solution:
Given equation of the circle is x² + y² – 6x – 4y = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -6, 2f = -4, c = 0
⇒ g = -3, f = -2, c = 0
The equation of a tangent to the circle x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
the equation of the tangent at (6, 4) is
x(6) + y(4) – 3(x + 6) – 2(y + 4) + 0 = 0
⇒ 6x + 4y – 3x – 18 – 2y – 8 = 0
⇒ 3x + 2y – 26 = 0

Alternate method:
Given equation of the circle is x² + y² – 6x – 4y = 0
x(x – 6) + y(y – 4) = 0, which is in diameter form where (0, 0) and (6, 4) are endpoints of diameter.

Slope of OP = \(\frac{4-0}{6-0}=\frac{2}{3}\)
Since, OP is perpendicular to the required tangent.
Slope of the required tangent = \(\frac{-3}{2}\)
the equation of the tangent at (6, 4) is
y – 4 = \(\frac{-3}{2}\) (x – 6)
⇒ 2(y – 4) = 3(x – 6)
⇒ 2y – 8 = -3x + 18
⇒ 3x + 2y – 26 = 0

Question 17.
Fihd the equation of tangent to circle x² + y² = 5, at the point (1, -2) on it.
Solution:
Given equation of the circle is x² + y² = 5
Comparing this equation with x² + y² = r², we get
r² = 5
The equation of a tangent to the circle x² + y² = r² at (x₁, y₁) is xx₁ + yy₁ = r²
the equation of the tangent at (1, -2) is
x(1) + y(-2) = 5
⇒ x – 2y = 5

Question 18.
Find the equation of tangent to circle x = 5 cos θ, y = 5 sin θ, at the point θ = \(\frac{\pi}{3}\) on it.
Solution:
The equation of a tangent to the circle x² + y² = r² at P(θ) is x cos θ + y sin θ = r
Here, r = 5, θ = \(\frac{\pi}{3}\)
the equation of the tangent at P(\(\frac{\pi}{3}\)) is
x cos \(\frac{\pi}{3}\) + y sin \(\frac{\pi}{3}\) = 5
⇒ \(x\left(\frac{1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=5\)
⇒ x + √3y = 10

Question 19.
Show that 2x + y + 6 = 0 is a tangent to x² + y² + 2x – 2y – 3 = 0. Find its point of contact.
Solution:
Given equation of circle is
x² + y² + 2x – 2y – 3 = 0 ….(i)
Given equation of line is 2x + y + 6 = 0
y = -6 – 2x ……(ii)
Substituting y = -6 – 2x in (i), we get
x + (-6 – 2x)² + 2x – 2(-6 – 2x) – 3 = 0
⇒ x² + 36 + 24x + 4x² + 2x + 12 + 4x – 3 = 0
⇒ 5x² + 30x + 45 = 0
⇒ x² + 6x + 9 = 0
⇒ (x + 3)² = 0
⇒ x = -3
Since, the roots are equal.
∴ 2x + y + 6 = 0 is a tangent to x² + y² + 2x – 2y – 3 = 0
Substituting x = -3 in (ii), we get
y = -6 – 2(-3) = -6 + 6 = 0
Point of contact = (-3, 0)

Question 20.
If the tangent at (3, -4) to the circle x² + y² = 25 touches the circle x² + y² + 8x – 4y + c = 0, find c.
Solution:
The equation of a tangent to the circle
x² + y² = r² at (x₁, y₁) is xx₁ + yy₁ = r²
Equation of the tangent at (3, -4) is
x(3) + y(-4) = 25
⇒ 3x – 4y – 25 = 0 ……(i)
Given equation of circle is x² + y² + 8x – 4y + c = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = 8, 2f = -4
⇒ g = 4, f = -2
∴ C = (-4, 2) and r = \(\sqrt{4^{2}+(-2)^{2}-c}=\sqrt{20-c}\)
Since line (i) is a tangent to this circle also, the perpendicular distance from C(-4, 2) to line (i) is equal to radius r.

Question 21.
Find the equations of the tangents to the circle x² + y² = 16 with slope -2.
Solution:
Given equation of the circle is x² + y² = 16
Comparing this equation with x² + y² = a², we get
a² = 16
Equations of the tangents to the circle x² + y² = a² with slope m are
\(y=m x \pm \sqrt{a^{2}\left(1+m^{2}\right)}\)
Here, m = -2, a² = 16
the required equations of the tangents are
y = \(-2 x \pm \sqrt{16\left[1+(-2)^{2}\right]}\)
⇒ y = \(-2 x \pm \sqrt{16(5)}\)
⇒ y = -2x ± 4√5
⇒ 2x + y ± 4√5 = 0

Question 22.
Find the equations of the tangents to the circle x² + y² = 4 which are parallel to 3x + 2y + 1 = 0.
Solution:
Given equation of the circle is x² + y² = 4
Comparing this equation with x² + y² = a², we get
a² = 4
Given equation of the line is 3x + 2y + 1 = 0
Slope of this line = \(\frac{-3}{2}\)
Since, the required tangents are parallel to the given line.
Slope of required tangents (m) = \(\frac{-3}{2}\)
Equations of the tangents to the circle x² + y² = a² with slope m are
y = mx ± \(\sqrt{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}\)
the required equations of the tangents are

Question 23.
Find the equations of the tangents to the circle x² + y² = 36 which are perpendicular to the line 5x + y = 2.
Solution:
Given equation of the circle is x² + y² = 36
Comparing this equaiton with x² + y² = a², we get
a² = 36
Given equation of line is 5x + y = 2
Slope of this line = -5
Since, the required tangents are perpendicular to the given line.
Slope of required tangents (m) = \(\frac{1}{5}\)
Equations of the tangents to the circle x² + y² = a² with slope m are
y = mx ± \(\sqrt{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}\)
the required equations of the tangents are

Question 24.
Find the equations of the tangents to the circle x² + y² – 2x + 8y – 23 = 0 having slope 3.
Solution:
Let the equation of the tangent with slope 3 be y = 3x + c.
3x – y + c = 0 ……(i)
Given equation of circle is x² + y² – 2x + 8y – 23 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -2, 2f = 8, c = -23
g = -1, f = 4, c = -23
The centre of the circle is C(1, -4)
and its radius = \(\sqrt{1+16+23}\)
= √40
= 2√10
Since line (i) is a tangent to this circle the perpendicular distance from C(1, -4) to line (i) is equal to radius r.
\(\left|\frac{3(1)+4+c}{\sqrt{9+1}}\right|\) = 2√10
⇒ \(\left|\frac{7+c}{\sqrt{10}}\right|\) = 2√10
⇒ (7 + c) = ± 20
⇒ 7 + c = 20 or 7 + c = -20
⇒ c = 13 or c = – 27
∴ Equations of the tangents are 3x – y + 13 = 0 and 3x – y – 21 = 0

Question 25.
Find the equation of the locus of a point, the tangents from which to the circle x² + y² = 9 are at right angles.
Solution:
Given equation of the circle is x² + y² = 9
Comparing this equation with x² + y² = a², we get
a² = 9
The locus of the point of intersection of perpendicular tangents is the director circle of the given circle.
The equation of the director circle of the circle x² + y² = a² is x² + y² = 2a².
the required equation is
x² + y² = 2(9)
x² + y² = 18

Alternate method:
Given equation of the circle is x² + y² = 9
Comparing this equation with x² + y² = a², we get a² = 9
Let P(x₁, y₁) be a point on the required locus.
Equations of the tangents to the circle x² + y² = a² with slope m are
y = mx ± \(\sqrt{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}\)
∴ Equations of the tangents are
y = mx ± \(\sqrt{9\left(\mathrm{~m}^{2}+1\right)}\)
⇒ y = mx ± 3\(\sqrt{1+m^{2}}\)
Since, these tangents pass through (x₁, y₁).
y₁ = mx₁ ± 3\(\sqrt{1+m^{2}}\)
⇒ y₁ – mx₁ = ± 3\(\sqrt{1+m^{2}}\)
⇒ (y₁ – mx₁)² = 9(1 + m²) ……[Squaring both the sides]
⇒ \(y_{1}^{2}-2 m x_{1} y_{1}+m^{2} x_{1}^{2}=9+9 m^{2}\)
⇒ \(\left(x_{1}^{2}-9\right) \mathrm{m}^{2}-2 \mathrm{~m} x_{1} y_{1}+\left(y_{1}^{2}-9\right)=0\)
This is a quadratic equation which has two roots m₁ and m₂.
m₁m₂ = \(\frac{y_{1}^{2}-9}{x_{1}^{2}-9}\)
Since, the tangents are at right angles.
m₁m₂ = -1
⇒ \(\frac{y_{1}^{2}-9}{x_{1}^{2}-9}=-1\)
⇒ \(y_{1}^{2}-9=9-x_{1}^{2}\)
⇒ \(x_{1}^{2}+y_{1}^{2}=18\)
Equation of the locus of point P is x² + y² = 18.

Question 26.
Tangents to the circle x² + y² = a² with inclinations, θ₁ and θ₂ intersect in P. Find the locus of P such that
(i) tan θ₁ + tan θ₂ = 0
(ii) cot θ₁ + cot θ₂ = 5
(iii) cot θ₁ . cot θ₂ = c
Solution:
Let P(x₁, y₁) be a point on the required locus.
Equations of the tangents to the circle x² + y² = a² with slope m are
y = mx ± \(\sqrt{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}\)
Since, these tangents pass through (x₁, y₁).

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Ex 6.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Ex 6.3

Question 1.
Write the parametric equations of the circles:
(i) x² + y² = 9
(ii) x² + y² + 2x – 4y – 4 = 0
(iii) (x – 3)² + (y + 4)² = 25
Solution:
(i) Given equation of the circle is
x² + y² = 9
⇒ x² + y² = 3²
Comparing this equation with x² + y² = r², we get r = 3
The parametric equations of the circle in terms of θ are
x = r cos θ and y = r sin θ
⇒ x = 3 cos θ and y = 3 sin θ

(ii) Given equation of the circle is
x² + y² + 2x – 4y – 4 = 0
⇒ x² + 2x + y² – 4y – 4 = 0
⇒ x² + 2x + 1 – 1 + y² – 4y + 4 – 4 – 4 = 0
⇒ (x² + 2x + 1 ) + (y² – 4y + 4) – 9 = 0
⇒ (x + 1)² + (y – 2)² = 9
⇒ (x + 1)² + (y – 2)² = 3²
Comparing this equation with (x – h)² + (y – k)² = r², we get
h = -1, k = 2 and r = 3
The parametric equations of the circle in terms of θ are
x = h + r cos θ and y = k + r sin θ
⇒ x = -1 + 3 cos θ and y = 2 + 3 sin θ

(iii) Given equation of the circle is
(x – 3)² + (y + 4)² = 25
⇒ (x – 3)² + (y + 4)² = 5²
Comparing this equation with (x – h)² + (y – k)² = r², we get
h = 3, k = -4 and r = 5
The parametric equations of the circle in terms of θ are
x = h + r cos θ and y = k + r sin θ
⇒ x = 3 + 5 cos θ and y = -4 + 5 sin θ

Question 2.
Find the parametric representation of the circle 3x² + 3y² – 4x + 6y – 4 = 0.
Solution:
Given equation of the circle is 3x² + 3y² – 4x + 6y – 4 = 0
Dividing throughout by 3, we get

Comparing this equation with (x – h)² + (y – k)² = r², we get
h = \(\frac{2}{3}\), k = -1 and r = \(\frac{5}{3}\)
The parametric representation of the circle in terms of θ are
x = h + r cos θ and y = k + r sin θ
⇒ x = \(\frac{2}{3}\) + \(\frac{5}{3}\) cos θ and y = -1 + \(\frac{5}{3}\) sin θ

Question 3.
Find the equation of a tangent to the circle x² + y² – 3x + 2y = 0 at the origin.
Solution:
Given equation of the circle is x² + y² – 3x + 2y = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -3, 2f = 2, c = 0
⇒ g = \(-\frac{3}{2}\), f = 1, c = 0
The equation of a tangent to the circle
x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is xx₁ +yy₁ + g(x + x₁) + f(y + y₁) + c = 0
The equation of the tangent at (0, 0) is
x(0) + y(0) + (\(-\frac{3}{2}\)) (x + 0) + 1(y + 0) + 0 = 0
⇒ \(-\frac{3}{2}\)x + y = 0
⇒ 3x – 2y = 0

Question 4.
Show that the line 7x – 3y – 1 = 0 touches the circle x² + y² + 5x – 7y + 4 = 0 at point (1, 2).
Solution:
Given equation of the circle is x² + y² + 5x – 7y + 4 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = 5, 2f = -7, c = 4
⇒ g = \(\frac{5}{2}\), f = \(\frac{-7}{2}\), c = 4
The equation of a tangent to the circle x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
The equation of the tangent at (1, 2) is

7x – 3y – 1 = 0, which is same as the given line.
The line 7x – 3y – 1 = 0 touches the given circle at (1, 2).

Question 5.
Find the equation of tangent to the circle x² + y² – 4x + 3y + 2 = 0 at the point (4, -2).
Solution:
Given equation of the circle is x² + y² – 4x + 3y + 2 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -4, 2f = 3, c = 2
g = -2, f = \(\frac{3}{2}\), c = 2
The equation of a tangent to the circle x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
The equation of the tangent at (4, -2) is
x(4) + y(-2) – 2(x + 4) + \(\frac{3}{2}\)(y – 2) + 2 = 0
⇒ 4x – 2y – 2x – 8 + \(\frac{3}{2}\) y – 3 + 2 = 0
⇒ 2x – \(\frac{1}{2}\)y – 9 = 0
⇒ 4x – y – 18 = 0

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Ex 6.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Ex 6.2

Question 1.
Find the centre and radius of each of the following circles:
(i) x² + y² – 2x + 4y – 4 = 0
(ii) x² + y² – 6x – 8y – 24 = 0
(iii) 4x² + 4y² – 24x – 8y – 24 = 0
Solution:
(i) Given equation of the circle is x² + y² – 2x + 4y – 4 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -2, 2f = 4 and c = -4
⇒ g = -1, f = 2 and c = -4
Centre of the circle = (-g, -f) = (1, -2)
and radius of the circle

(ii) Given equation of the circle is x² + y² – 6x – 8y – 24 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -6, 2f = -8 and c = -24
⇒ g = -3, f = -4 and c = -24
Centre of the circle = (-g, -f) = (3, 4)
and radius of the circle

(iii) Given equation of the circle is 4x² + 4y² – 24x – 8y – 24 = 0
Dividing throughout by 4, we get x² + y² – 6x – 2y – 6 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -6, 2f = -2 and c = -6
⇒ g = -3, f = -1 and c = -6
Centre of the circle = (-g, -f) = (3, 1)
and radius of the circle

Question 2.
Show that the equation 3x² + 3y² + 12x + 18y – 11 = 0 represents a circle.
Solution:
Given equation is 3x² + 3y² + 12x + 18y – 11 = 0
Dividing throughout by 3, we get
x² + y² + 4x + 6y – \(\frac{11}{3}\) = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = 4, 2f = 6, c = \(\frac{-11}{3}\)
⇒ g = 2, f = 3, c = \(\frac{-11}{3}\)
Now, g² + f² – c = (2)² + (3)² – (\(\frac{-11}{3}\))
= 4 + 9 + \(\frac{11}{3}\)
= \(\frac{50}{3}\) > 0
∴ The given equation represents a circle.

Question 3.
Find the equation of the circle passing through the points (5, 7), (6, 6), and (2, -2).
Solution:
Let C(h, k) be the centre of the required circle.
Since the required circle passes through points A(5, 7), B(6, 6), and D(2, -2),
CA = CB = CD = radius

Consider, CA = CD
By distance formula,
\(\sqrt{(\mathrm{h}-5)^{2}+(\mathrm{k}-7)^{2}}=\sqrt{(\mathrm{h}-2)^{2}+[\mathrm{k}-(-2)]^{2}}\)
Squaring both the sides, we get
⇒ (h – 5)² + (k – 7)² = (h – 2)² + (k + 2)²
⇒ h² – 10h + 25 + k² – 14k + 49 = h² – 4h + 4 + k² + 4k + 4
⇒ -10h – 14k + 74 = -4h + 4k + 8
⇒ 6h + 18k – 66 = 0
⇒ h + 3k – 11 = 0 …..(i)
Consider, CB = CD
By distance formula,
\(\sqrt{(h-6)^{2}+(k-6)^{2}}=\sqrt{(h-2)^{2}+[k-(-2)]^{2}}\)
Squaring both the sides, we get
⇒ (h – 6)² + (k – 6)² = (h – 2)² + (k + 2)²
⇒ h² – 12h + 36 + k² – 12k + 36 = h² – 4h + 4 + k² + 4k + 4
⇒ -12h – 12k + 72 = -4h + 4k + 8
⇒ 8h + 16k – 64 = 0
⇒ h + 2k – 8 = 0 ……(ii)
By (i) – (ii), we get k = 3
Substituting k = 3 in (i), we get
h + 3(3) – 11 = 0
⇒ h + 9 – 11 = 0
⇒ h = 2
Centre of the circle is C(2, 3).
radius (r) = CD
= \(\sqrt{(2-2)^{2}+(3+2)^{2}}\)
= \(\sqrt{0+5^{2}}\)
= √25
= 5
The equation of a circle with centre at (h, k) and radius r is given by (x – h)²+ (y – k)² = r²
Here, h = 2, k = 3
The required equation of the circle is
(x – 2)² + (y – 3)² = 5²
⇒ x² – 4x + 4 + y² – 6y + 9 = 25
⇒ x² + y² – 4x – 6y + 4 + 9 – 25 = 0
⇒ x² + y² – 4x – 6y – 12 = 0

Question 4.
Show that the points (3, -2), (1, 0), (-1, -2) and (1, -4) are concyclic.
Solution:
Let the equation of the circle passing through the points (3, -2), (1, 0) and (-1, -2) be
x² + y² + 2gx + 2fy + c = 0 …..(i)
For point (3, -2),
Substituting x = 3 and y = -2 in (i), we get
9 + 4 + 6g – 4f + c = 0
⇒ 6g – 4f + c = -13 ….(ii)
For point (1, 0),
Substituting x = 1 andy = 0 in (i), we get
1 + 0 + 2g + 0 + c = 0
⇒ 2g + c = -1 ……(iii)
For point (-1, -2),
Substituting x = -1 and y = -2, we get
1 + 4 – 2g – 4f + c = 0
⇒ 2g + 4f – c = 5 …….(iv)
Adding (ii) and (iv), we get
8g = -8
⇒ g = -1
Substituting g = -1 in (iii), we get
-2 + c = -1
⇒ c = 1
Substituting g = -1 and c = 1 in (iv), we get
-2 + 4f – 1 = 5
⇒ 4f = 8
⇒ f = 2
Substituting g = -1, f = 2 and c = 1 in (i), we get
x² + y² – 2x + 4y + 1 = 0 ……….(v)
If (1, -4) satisfies equation (v), the four points are concyclic.
Substituting x = 1, y = -4 in L.H.S of (v), we get
L.H.S. = (1)² + (-4)² – 2(1) + 4(-4) + 1
= 1 + 16 – 2 – 16 + 1
= 0
= R.H.S.
Point (1, -4) satisfies equation (v).
∴ The given points are concyclic.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Ex 6.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Ex 6.1

Question 1.
Find the equation of a circle with
(i) centre at origin and radius 4.
(ii) centre at (-3, -2) and radius 6.
(iii) centre at (2, -3) and radius 5.
(iv) centre at (-3, -3) passing through point (-3, -6).
Solution:
(i) The equation of a circle with centre at origin and radius ‘r’ is given by
x² + y² = r²
Here, r = 4
∴ The required equation of the circle is x² + y² = 4² i.e., x² + y² = 16.

(ii) The equation of a circle with centre at (h, k) and radius ‘r’ is given by
(x – h)² + (y – k)² = r²
Here, h = -3, k = -2 and r = 6
∴ The required equation of the circle is
[x – (-3)]² + [y – (-2)]² = 6²
⇒ (x + 3)² + (y + 2)² = 36
⇒ x² + 6x + 9 + y² + 4y + 4 – 36 = 0
⇒ x² + y² + 6x + 4y – 23 = 0

(iii) The equation of a circle with centre at (h, k) and radius ‘r’ is given by
(x – h)² + (y – k)² = r²
Here, h = 2, k = -3 and r = 5
The required equation of the circle is
(x – 2)² + [y – (-3)]² = 5²
⇒ (x – 2)² + (y + 3)² = 25
⇒ x² – 4x + 4 + y² + 6y + 9 – 25 = 0
⇒ x² + y² – 4x + 6y – 12 = 0

(iv) Centre of the circle is C (-3, -3) and it passes through the point P (-3, -6).

The equation of a circle with centre at (h, k) and radius ‘r’ is given by
(x – h)² + (y – k)² = r²
Here, h = -3, k = -3, r = 3
The required equation of the circle is
[x – (-3)]² + [y – (-3)]² = 3²
⇒ (x + 3)² + (y + 3)² = 9
⇒ x² + 6x + 9 + y² + 6y + 9 – 9 = 0
⇒ x² + y² + 6x + 6y + 9 = 0

Check:
If the point (-3, -6) satisfies x² + y² + 6x + 6y + 9 = 0, then our answer is correct.
L.H.S. = x² + y² + 6x + 6y + 9
= (-3)² + (-6)² + 6(-3) – 6(-6) + 9
= 9 + 36 – 18 – 36 + 9
= 0
= R.H.S.
Thus, our answer is correct.

Question 2.
Find the centre and radius of the following circles:
(i) x² + y² = 25
(ii) (x – 5)² + (y – 3)² = 20
(iii) \(\left(x-\frac{1}{2}\right)^{2}+\left(y+\frac{1}{3}\right)^{2}=\frac{1}{36}\)
Solution:
(i) Given equation of the circle is
x² + y² = 25
⇒ x² + y² = (5)²
Comparing this equation with x² + y² = r², we get r = 5
Centre of the circle is (0, 0) and radius of the circle is 5.

(ii) Given equation of the circle is
(x – 5)² + (y – 3)² = 20
⇒ (x – 5)² + (y – 3)² = (√20)²
Comparing this equation with (x – h)² + (y – k)² = r², we get
h = 5, k = 3 and r = √20 = 2√5
Centre of the circle = (h, k) = (5, 3)
and radius of the circle = 2√5.

(iii) Given the equation of the circle is

Comparing this equation with (x – h)² + (y – k)² = r², we get
h = \(\frac{1}{2}\), k = \(\frac{-1}{3}\) and r = \(\frac{1}{6}\)
Centre of the circle = (h, k) = (\(\frac{1}{2}\), \(\frac{-1}{3}\)) and radius of the circle = \(\frac{1}{6}\)

Question 3.
Find the equation of the circle with centre
(i) at (a, b) and touching the Y-axis.
(ii) at (-2, 3) and touching the X-axis.
(iii) on the X-axis and passing through the origin having radius 4.
(iv) at (3, 1) and touching the line 8x – 15y + 25 = 0.
Solution:
(i) Since the circle is touching the Y-axis, the radius of the circle is X-co-ordinate of the centre.

∴ r = a
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = a, k = b
The required equation of the circle is
⇒ (x – a)² + (y – b)² = a²
⇒ x² – 2ax + a² + y² – 2by + b² = a²
⇒ x² + y² – 2ax – 2by + b² = 0

(ii) Since the circle is touching the X-axis, the radius of the circle is the Y co-ordinate of the centre.

∴ r = 3
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = -2, k = 3
The required equation of the circle is
⇒ (x + 2)² + (y – 3)² = 3²
⇒ x² + 4x + 4 + y² – 6y + 9 = 9
⇒ x² + y² + 4x – 6y + 4 = 0

(iii) Let the co-ordinates of the centre of the required circle be C (h, 0).
Since the circle passes through the origin i.e., O(0, 0)
OC = radius
⇒ \(\sqrt{(h-0)^{2}+(0-0)^{2}}=4\)
⇒ h² = 16
⇒ h = ±4

the co-ordinates of the centre are (4, 0) or (-4, 0).
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = ± 4, k = 0, r = 4
The required equation of the circle is
⇒ (x – 4)² + (y – 0)² = 4² or (x + 4)² + (y – 0)² = 4²
⇒ x² – 8x + 16 + y² = 16 or x² + 8x + 16 + y² = 16
⇒ x² + y² – 8x = 0 or x² + y² + 8x = 0

(iv) Centre of the circle is C (3, 1).
Let the circle touch the line 8x – 15y + 25 = 0 at point M.

CM = radius (r)
CM = Length of perpendicular from centre C(3, 1) on the line 8x – 15y + 25 = 0

The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = 3, k = 1 and r = 2
The required equation of the circle is
⇒ (x – 3)² + (y – 1)² = 2²
⇒ x² – 6x + 9 + y² – 2y + 1 = 4
⇒ x² + y² – 6x – 2y + 10 – 4 = 0
⇒ x² + y² – 6x – 2y + 6 = 0

Question 4.
Find the equation of the circle, if the equations of two diameters are 2x + y = 6 and 3x + 2y = 4 and radius is 9.
Solution:

Given equations of diameters are 2x + y = 6 and 3x + 2y = 4.
Let C (h, k) be the centre of the required circle.
Since point of intersection of diameters is the centre of the circle,
x = h, y = k
Equations of diameters become
2h + k = 6 …..(i)
and 3h + 2k = 4 ……..(ii)
By (ii) – 2 × (i), we get
-h = -8
⇒ h = 8
Substituting h = 8 in (i), we get
2(8) + k = 6
⇒ k = 6 – 16
⇒ k = -10
Centre of the circle is C (8, -10) and radius, r = 9
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = 8, k = -10
The required equation of the circle is
⇒ (x – 8)² + (y + 10)² = 9²
⇒ x² – 16x + 64 + y² + 20y + 100 = 81
⇒ x² + y² – 16x + 20y + 100 + 64 – 81 = 0
⇒ x² + y² – 16x + 20y + 83 = 0

Question 5.
If y = 2x is a chord of the circle x² + y² – 10x = 0, find the equation of the circle with this chord as diameter.
Solution:

y = 2x is the chord of the given circle.
It satisfies the equation of a given circle.
Substituting y = 2x in x² + y² – 10x = 0, we get
⇒ x² + (2x)² – 10x = 0
⇒ x² + 4x² – 10x = 0
⇒ 5x² – 10x = 0
⇒ 5x(x – 2) = 0
⇒ x = 0 or x = 2
When x = 0, y = 2x = 2(0) = 0
∴ A = (0, 0)
When x = 2, y = 2x = 2 (2) = 4
∴ B = (2, 4)
End points of chord AB are A(0, 0) and B(2, 4).
Chord AB is the diameter of the required circle.
The equation of a circle having (x₁, y₁) and (x₂, y₂) as end points of diameter is given by
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
Here, x₁ = 0, y₁ = 0, x₂ = 2, y₂ = 4
The required equation of the circle is
⇒ (x – 0) (x – 2) + (y – 0) (y – 4 ) = 0
⇒ x² – 2x + y² – 4y = 0
⇒ x² + y² – 2x – 4y = 0

Question 6.
Find the equation of a circle with a radius of 4 units and touch both the co-ordinate axes having centre in the third quadrant. Solution:
The radius of the circle = 4 units
Since the circle touches both the co-ordinate axes and its centre is in the third quadrant,
the centre of the circle is C(-4, -4).

The equation of a circle with centre at (h, k) and radius r is given by (x – h)² + (y – k)² = r²
Here, h = -4, k = -4, r = 4
the required equation of the circle is
⇒ [x – (-4)]² + [y – (-4)]² = 4²
⇒ (x + 4)² + (y + 4)² = 16
⇒ x² + 8x + 16 + y² + 8y + 16 – 16 = 0
⇒ x² + y² + 8x + 8y + 16 = 0

Question 7.
Find the equation of the circle passing through the origin and having intercepts 4 and -5 on the co-ordinate axes.
Solution:
Let the circle intersect X-axis at point A and intersect Y-axis at point B.
the co-ordinates of point A are (4, 0) and the co-ordinates of point B are (0, -5).

Since ∠AOB is a right angle,
AB represents the diameter of the circle.
The equation of a circle having (x₁, y₁) and (x₂, y₂) as end points of diameter is given by
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
Here, x₁ = 4, y₁ = 0, x₂ = 0, y₂ = -5
The required equation of the circle is
⇒ (x – 4) (x – 0) + (y – 0) [y – (-5)] = 0
⇒ x(x – 4) + y(y + 5) = 0
⇒ x² – 4x + y² + 5y = 0
⇒ x² + y² – 4x + 5y = 0

Question 8.
Find the equation of a circle passing through the points (1, -4), (5, 2) and having its centre on line x – 2y + 9 = 0.
Solution:

Let C(h, k) be the centre of the required circle which lies on the line x – 2y + 9 = 0.
Equation of line becomes
h – 2k + 9 = 0 …..(i)
Also, the required circle passes through points A(1, -4) and B(5, 2).
CA = CB = radius
CA = CB
By distance formula,
\(\sqrt{(\mathrm{h}-1)^{2}+[\mathrm{k}-(-4)]^{2}}=\sqrt{(\mathrm{h}-5)^{2}+(\mathrm{k}-2)^{2}}\)
Squaring both the sides, we get
⇒ (h – 1)² + (k + 4)² = (h – 5)² + (k – 2)²
⇒ h² – 2h + 1 + k² + 8k + 16 = h² – 10h + 25 + k² – 4k + 4
⇒ -2h + 8k + 17 = -10h – 4k + 29
⇒ 8h + 12k – 12 = 0
⇒ 2h + 3k – 3 = 0 ……(ii)
By (ii) – (i) × 2, we get
7k = 21
⇒ k = 3
Substituting k = 3 in (i), we get
h – 2(3) + 9 = 0
⇒ h – 6 + 9 = 0
⇒ h = -3
Centre of the circle is C(-3, 3).
radius (r) = CA

The equation of a circle with centre at (h, k) and radius r is given by (x – h)² + (y – k)² = r²
Here, h = -3, k = 3, r = √65
The required equation of the circle is
⇒ [x – (-3)]² + (y – 3)² = (√65)²
⇒ (x + 3)² + (y – 3)² = 65
⇒ x² + 6x + 9 + y² – 6y + 9 – 65 = 0
⇒ x² + y² + 6x – 6y – 47 = 0

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Miscellaneous Exercise 5

(I) Select the correct option from the given alternatives.

Question 1.
If A is (5, -3) and B is a point on the X-axis such that the slope of line AB is -2, then B ≡
(a) (7, 2)
(b) (\(\frac{7}{2}\), 0)
(c) (0, \(\frac{7}{2}\))
(d) (\(\frac{2}{7}\), 0)
Answer:
(b) (\(\frac{7}{2}\), 0)
Hint:
Let B(x, 0) be the point on X-axis.
We have A = (5, -3)
slope of AB = -2
⇒ \(\frac{0-(-3)}{x-5}\) = -2
⇒ 3 = -2(x – 5)
⇒ 3 = -2x + 10
⇒ x = \(\frac{7}{2}\)
Co-ordinates of point B = (\(\frac{7}{2}\), 0)

Question 2.
If the point (1, 1) lies on the line passing through the points (a, 0) and (0, b), then \(\frac{1}{a}+\frac{1}{b}=\)
(a) -1
(b) 0
(c) 1
(d) \(\frac{1}{a b}\)
Answer:
(c) 1
Hint:
Line passes through (a, 0), (0, b).
x-intercept = a, y-intercept = b
∴ Equation of line is \(\frac{x}{a}+\frac{y}{b}=1\) …….(i)
Since line (i) passes through (1, 1), (1, 1) satisfies (i)
∴ \(\frac{1}{a}+\frac{1}{b}=1\)

Question 3.
If A(1, -2), B(-2, 3) and C(2, -5) are the vertices of ΔABC, then the equation of median BE is
(a) 7x + 13y + 47 = 0
(b) 13x + 7y + 5 = 0
(c) 7x – 13y + 5 = 0
(d) 13x – 7y – 5 = 0
Answer:
(b) 13x + 7y + 5 = 0
Hint:

Question 4.
The equation of the line through (1, 2), which makes equal intercepts on the axes, is
(a) x + y = 1
(b) x + y = 2
(c) x + y = 4
(d) x + y = 3
Answer:
(d) x + y = 3
Hint:
Let the equation of required line be
\(\frac{x}{a}+\frac{y}{b}=1\) ……..(i)
Since the line makes equal intercepts on the axes, a = b
\(\frac{x}{a}+\frac{y}{a}=1\)
∴ x + y = a ……(ii)
But, equation (ii) passes through (1, 2).
1 + 2 = a
∴ a = 3
Substituting a = 3 in equation (ii), we get
x + y = 3

Question 5.
If the line kx + 4y = 6 passes through the point of intersection of the two lines 2x + 3y = 4 and 3x + 4y = 5, then k =
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2
Hint:
Given two lines are
2x + 3y = 4 ……(i)
3x + 4y = 5 …….(ii)
Multiplying (i) by 3 and (ii) by 2 and then subtracting, we get
y = 2
Substituting y = 2 in (i), we get
x = -1
∴ Point of intersection of lines (i) and (ii) is (-1, 2).
Given that the line kx + 4y = 6 passes through (-1, 2).
k(-1) + 4(2) = 6
∴ k = 2

Question 6.
The equation of a line, having inclination 120° with positive direction of X-axis, which is at a distance of 3 units from the origin is
(a) √3x ± y + 6 = 0
(b) √3x + y ± 6 = 0
(c) x + y = 6
(d) x + y = -6
Answer:
(b) √3x + y ± 6 = 0
Hint:

Here, α = 30° and p = 3 units
Equation of line with inclination a and distance from origin as p is
x cos α + y sin α = p
∴ x cos 30° + y sin 30° = ±3
∴ \(\frac{\sqrt{3} x}{2}+\frac{y}{2}=\pm 3\)
∴ √3x + y ± 6 = 0

Question 7.
A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is
(a) \(\frac{1}{3}\)
(b) \(\frac{2}{3}\)
(c) 1
(d) \(\frac{4}{3}\)
Answer:
(d) \(\frac{4}{3}\)
Hint:
Slope of line 3x + y = 3 is -3
∴ Slope of line perpendicular to given line = \(\frac{1}{3}\)
Equation of required line passing through (2, 2) and having slope \(\frac{1}{3}\) is
y – 2 = \(\frac{1}{3}\)(x – 2)
3y – 6 = x – 2
∴ x – 3y + 4 = 0
∴ y-intercept = \(\frac{-4}{-3}=\frac{4}{3}\)

Question 8.
The angle between the line √3x – y – 2 = 0 and x – √3y + 1 = 0 is
(a) 15°
(b) 30°
(c) 45°
(d) 60°
Answer:
(b) 30°
Hint:

Question 9.
If kx + 2y – 1 = 0 and 6x – 4y + 2 = 0 are identical lines, then determine k.
(a) -3
(b) \(-\frac{1}{3}\)
(c) \(\frac{1}{3}\)
(d) 3
Answer:
(a) -3
Hint:
Lines kx + 2y – 1 = 0 and 6x – 4y + 2 = 0 are identical.
∴ \(\frac{k}{6}=\frac{2}{-4}=\frac{-1}{2}\)
∴ k = -3

Question 10.
Distance between the two parallel lines y = 2x + 7 and y = 2x + 5 is
(a) \(\frac{\sqrt{2}}{\sqrt{5}}\)
(b) \(\frac{1}{\sqrt{5}}\)
(c) \(\frac{\sqrt{5}}{2}\)
(d) \(\frac{2}{\sqrt{5}}\)
Answer:
(d) \(\frac{2}{\sqrt{5}}\)
Hint:
Here, c₁ = 7, c₂ = 5, a = 2 and b = -1
Distance between parallel lines

II. Answer the following questions.

Question 1.
Find the value of k:
(a) if the slope of the line passing through the points P(3, 4), Q(5, k) is 9.
(b) the points A(1, 3), B(4, 1), C(3, k) are collinear.
(c) the point P(1, k) lies on the line passing through the points A(2, 2) and B(3, 3).
Solution:
(a) Given, P(3, 4), Q(5, k) and
Slope of PQ = 9
\(\frac{\mathrm{k}-4}{5-3}\) = 9
\(\frac{\mathrm{k}-4}{2}\) = 9
k – 4 = 18
k = 22

(b) Given, points A(1, 3), B(4, 1) and C(3, k) are collinear.
Slope of AB = Slope of BC
\(\frac{1-3}{4-1}=\frac{k-1}{3-4}\)
\(\frac{-2}{3}=\frac{\mathrm{k}-1}{-1}\)
2 = 3k – 3
k = \(\frac{5}{3}\)

(c) Given, point P(1, k) lies on the line joining A(2, 2) and B(3, 3).
Slope of AB = Slope of BP
\(\frac{3-2}{3-2}=\frac{3-k}{3-1}\)
1 = \(\frac{3-\mathrm{k}}{2}\)
2 = 3 – k
k = 1

Question 2.
Reduce the equation 6x + 3y + 8 = 0 into slope-intercept form. Hence, find its slope.
Solution:
Given equation is 6x + 3y + 8 = 0, which can be written as
3y = – 6x – 8
y = \(\frac{-6 x}{3}-\frac{8}{3}\)
y = -2x – \(\frac{8}{3}\)
This is of the form y = mx + c with m = -2
y = -2x – \(\frac{8}{3}\) is in slope-intercept form with slope = -2

Question 3.
Find the distance of the origin from the line x = -2.
Solution:
Given equation of line is x = -2
This equation represents a line parallel to Y-axis and at a distance of 2 units to the left of Y-axis.
∴ Distance of the origin from the line is 2 units.

Question 4.
Does point A(2, 3) lie on the line 3x + 2y – 6 = 0? Give reason.
Solution:
Given equation is 3x + 2y – 6 = 0.
Substituting x = 2 and y = 3 in L.H.S. of given equation, we get
L.H.S. = 3x + 2y – 6
= 3(2) + 2(3) – 6
= 6
≠ R.H.S.
∴ Point A does not lie on the given line.

Question 5.
Which of the following lines passes through the origin?
(a) x = 2
(b) y = 3
(c) y = x + 2
(d) 2x – y = 0
Answer:
(d) 2x – y = 0
Hint:
Any line passing through origin is of the form y = mx or ax + by = 0.
Here in the given option, 2x – y = 0 is in the form ax + by = 0.
∴ Option (d) is the correct answer.

Question 6.
Obtain the equation of the line which is:
(a) parallel to the X-axis and 3 units below it.
(b) parallel to the Y-axis and 2 units to the left of it.
(c) parallel to the X-axis and making an intercept of 5 on the Y-axis.
(d) parallel to the Y-axis and making an intercept of 3 on the X-axis.
Solution:
(a) Equation of a line parallel to X-axis is y = k.
Since the line is at a distance of 3 units below X-axis, k = -3
∴ The equation of the required line is y = -3.

(b) Equation of a line parallel to Y-axis is x = h.
Since the line is at a distance of 2 units to the left of Y-axis, h = -2
∴ The equation of the required line is x = -2.

(c) Equation of a line parallel to X-axis with y-intercept ‘k’ is y = k.
Here, y-intercept = 5
∴ The equation of the required line is y = 5.

(d) Equation of a line parallel to Y-axis with x-intercept ‘h’ is x = h.
Here, x-intercept = 3
∴ The equation of the required line is x = 3.

Question 7.
Obtain the equation of the line containing the point:
(i) (2, 3) and parallel to the X-axis.
(ii) (2, 4) and perpendicular to the Y-axis.
Solution:
(i) Equation of a line parallel to X-axis is of the form y = k.
Since the line passes through (2, 3), k = 3
∴ The equation of the required line is y = 3.

(ii) Equation of a line perpendicular to Y-axis
i.e., parallel to X-axis, is of the form y = k.
Since the line passes through (2, 4), k = 4
∴ The equation of the required line is y = 4.

Question 8.
Find the equation of the line:
(a) having slope 5 and containing point A(-1, 2).
(b) containing the point T(7, 3) and having inclination 90°.
(c) through the origin which bisects the portion of the line 3x + 2y = 2 intercepted between the co-ordinate axes.
Solution:
(a) Given, slope(m) = 5 and the line passes through A(-1, 2).
Equation of the line in slope point form is y – y1 = m(x – x1)
The equation of the required line is
y – 2 = 5(x + 1)
y – 2 = 5x + 5
∴ 5x – y + 7 = 0

(b) Given, Inclination of line = θ = 90°
the required line is parallel to Y-axis.
Equation of a line parallel to Y-axis is of the form x = h.
Since the line passes through (7, 3), h = 7
∴ The equation of the required line is x = 7.

(c) Given equation of the line is 3x + 2y = 2.

\(\frac{3 x}{2}+\frac{2 y}{2}=1\)
\(\frac{x}{\frac{2}{3}}+\frac{y}{1}=1\)
This equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\), with a = \(\frac{2}{3}\), b= 1.
The line 3x + 2y = 2 intersects the X-axis at A(\(\frac{2}{3}\), 0) and Y-axis at B(0, 1).
Required line is passing through the midpoint of AB.
Midpoint of AB = \(\left(\frac{\frac{2}{3}+0}{2}, \frac{0+1}{2}\right)=\left(\frac{1}{3}, \frac{1}{2}\right)\)
∴ Required line passes through (0, 0) and \(\left(\frac{1}{3}, \frac{1}{2}\right)\).
Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the required line is
\(\frac{y-0}{\frac{1}{2}-0}=\frac{x-0}{\frac{1}{3}-0}\)
2y = 3x
∴ 3x – 2y = 0

Question 9.
Find the equation of the line passing through the points S(2, 1) and T(2, 3).
Solution:
The required line passes through the points S(2, 1) and T(2, 3).
Since both the given points have same x co-ordinates i.e. 2
the given points lie on a line parallel to Y-axis.
∴ The equation of the required line is x = 2.

Question 10.
Find the distance of the origin from the line 12x + 5y + 78 = 0.
Solution:
Let p be the perpendicular distance of origin from the line 12x + 5y + 78 = 0.
Here, a = 12, b = 5, c = 78

Question 11.
Find the distance between the parallel lines 3x + 4y + 3 = 0 and 3x + 4y + 15 = 0.
Solution:
Equations of the given parallel lines are 3x + 4y + 3 = 0 and 3x + 4y + 15 = 0
Here, a = 3, b = 4, c₁ = 3 and c₂ = 15
∴ Distance between the parallel lines

Question 12.
Find the equation of the line which contains the point A(3, 5) and makes equal intercepts on the co-ordinates axes.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}=1\) …….(i)
This line passes through A(3, 5).
∴ \(\frac{3}{a}+\frac{5}{b}=1\) ……..(ii)
Since the required line makes equal intercepts on the co-ordinates axes,
a = b …….(iii)
Substituting the value of b in (ii), we get
\(\frac{3}{a}+\frac{5}{a}=1\)
∴ a = 8
∴ b = 8 …… [From (iii)]
Substituting the values of a and b in equation (i), the equation of the required line is
\(\frac{x}{8}+\frac{y}{8}=1\)
∴ x + y = 8

Case II: Line passing through origin.
Slope of line passing through origin and A(3, 5) is
m = \(\frac{5-0}{3-0}=\frac{5}{3}\)
∴ Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is
y = \(\frac{5}{3}\)x
∴ 5x – 3y = 0

Question 13.
The vertices of a triangle are A(1, 4), B(2, 3) and C(1, 6). Find equations of
(a) the sides
(b) the medians
(c) perpendicular bisectors of sides
(d) altitudes of ?ABC
Solution:
Vertices of ∆ABC are A(1, 4), B(2, 3) and C(1, 6)
(a) Equation of the line in two point form is \(\frac{y-y_{1}}{y_{2}-y_{1}}\) = \(\frac{x-x_{1}}{x_{2}-x_{1}}\)
Equation of side AB is
\(\frac{y-4}{3-4}=\frac{x-1}{2-1}\)
y – 4 = -1(x – 1)
y – 4 = -x + 1
x + y = 5
Equation of side BC is
\(\frac{y-3}{6-3}=\frac{x-2}{1-2}\)
-1(y – 3) = 3(x – 2)
-y + 3 = 3x – 6
∴ 3x + y = 9
Since both the points A and C have same x co-ordinates i.e. 1
the points A and C lie on a line parallel to Y-axis.
∴ The equation of side AC is x = 1.

(b) Let D, E and F be the midpoints of sides AC and AB respectively of ∆ABC.

(c) Slope of side BC = \(\left(\frac{6-3}{1-2}\right)=\left(\frac{3}{-1}\right)\) = -3
Slope of perpendicular bisector of BC is \(\frac{1}{3}\) and the line passes through \(\left(\frac{3}{2}, \frac{9}{2}\right)\).
Equation of the perpendicular bisector of side BC is
\(\left(y-\frac{9}{2}\right)=\frac{1}{3}\left(x-\frac{3}{2}\right)\)
3(2y – 9) = (2x – 3)
6y – 27 = 2x – 3
2x – 6y + 24 = 0
∴ x – 3y + 12 = 0
Since both the points A and C have same x co-ordinates i.e. 1
the points A and C lie on the line x = 1.
AC is parallel to Y-axis and therefore, perpendicular bisector of side AC is parallel to X-axis.
Since, the perpendicular bisector of side AC passes through E(1, 5).
The equation of perpendicular bisector of side AC is y = 5.
Slope of side AB = \(\left(\frac{3-4}{2-1}\right)\) = -1
Slope of perpendicular bisector of AB is 1 and the line passes through \(\left(\frac{3}{2}, \frac{7}{2}\right)\).
Equation of the perpendicular bisector of side AB is
\(\left(y-\frac{7}{2}\right)=1\left(x-\frac{3}{2}\right)\)
2y – 7 = 2x – 3
2x – 2y + 4 = 0
∴ x – y + 2 = 0

(d) Let AX, BY, and CZ be the altitudes through the vertices A, B and C respectively of ∆ABC.

Slope of BC = -3
Slope of AX = \(\frac{1}{3}\) ……[∵ AX ⊥ BC]
Since altitude AX passes through (1, 4) and has slope \(\frac{1}{3}\),
equation of altitude AX is
y – 4 = \(\frac{1}{3}\)(x – 1)
3y – 12 = x – 1
∴ x – 3y + 11 = 0
Since both the points A and C have same x co-ordinates i.e. 1
the points A and C lie on the line x = 1.
AC is parallel to Y-axis and therefore, altitude BY is parallel to X-axis.
Since the altitude BY passes through B(2, 3), the equation of altitude BY is y = 3.
Also, slope of AB = -1
Slope of CZ = 1
Since altitude CZ passes through (1, 6) and has slope 1,
equation of altitude CZ is
y – 6 = 1(x – 1)
∴ x – y + 5 = 0

Question 14.
Find the equation of the line which passes through the point of intersection of lines x + y – 3 = 0, 2x – y + 1 = 0 and which is parallel to X-axis.
Solution:
Let u ≡ x + y – 3 = 0 and v ≡ 2x – y + 1 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
(x + y – 3) + k(2x – y + 1) = 0 …..(i)
x + y – 3 + 2kx – ky + k = 0
x + 2kx + y – ky – 3 + k = 0
(1 + 2k)x + (1 – k)y – 3 + k = 0
But, this line is parallel to X-axis
Its slope = 0
⇒ \(\frac{-(1+2 k)}{1-k}=0\)
⇒ 1 + 2k = 0
⇒ k = \(\frac{-1}{2}\)
Substituting the value of k in (i), we get
(x + y – 3) + \(\frac{-1}{2}\) (2x – y + 1) = 0
⇒ 2(x + y – 3) – (2x – y + 1 ) = 0
⇒ 2x + 2y – 6 – 2x + y – 1 = 0
⇒ 3y – 7 = 0, which is the equation of the required line.

Question 15.
Find the equation of the line which passes through the point of intersection of lines x + y + 9 = 0, 2x + 3y + 1 = 0 and which makes x-intercept 1.
Solution:
Let u ≡ x + y + 9 = 0 and v ≡ 2x + 3y + 1 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
(x + y + 9) + k(2x + 3y + 1) = 0 ……(i)
⇒ x + y + 9 + 2kx + 3ky + k = 0
⇒ (1 + 2k)x + (1 + 3k)y + 9 + k = 0
But, x-intercept of this line is 1.
⇒ \(\frac{-(9+\mathrm{k})}{1+2 \mathrm{k}}\)
⇒ -9 – k = 1 + 2k
⇒ k = \(\frac{-10}{3}\)
Substituting the value of k in (i), we get
(x + y + 9) + (\(\frac{-10}{3}\)) (2x + 3y + 1) = 0
⇒ 3(x + y + 9) – 10(2x + 3y + 1) = 0
⇒ 3x + 3y + 27 – 20x – 30y – 10 = 0
⇒ -17x – 27y+ 17 = 0
⇒ 17x + 27y – 17 = 0, which is the equation of the required line.

Question 16.
Find the equation of the line through A(-2, 3) and perpendicular to the line through S(1, 2) and T(2, 5).
Solution:
Slope of ST = \(\frac{5-2}{2-1}\) = 3
Since the required line is perpendicular to ST,
slope of required line = \(\frac{-1}{3}\) and line passes through A(-2, 3)
Equation of the line in slope point form is y – y₁ = m(x – x₁)
The equation of the required line is
y – 3 = \(\frac{-1}{3}\)(x + 2)
⇒ 3(y – 3) = -(x + 2)
⇒ 3y – 9 = -x – 2
⇒ x + 3y = 7

Question 17.
Find the x-intercept of the line whose slope is 3 and which makes intercept 4 on the Y-axis.
Solution:
Equation of a line having slope ‘m’ and y-intercept ‘c’ is y = mx + c
Given, m = 3, c = 4
The equation of the line is y = 3x + 4
3x – y = -4
\(\frac{3 x}{(-4)}-\frac{y}{(-4)}=1\)
\(\frac{x}{\left(\frac{-4}{3}\right)}+\frac{y}{4}=1\)
This equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\), where
x-intercept = a
x-intercept = \(\frac{-4}{3}\)

Alternate Method:

Let θ be the inclination of the line.
Then tan θ = 3 …..[∵ slope = 3 (given)]
\(\frac{\mathrm{OB}}{\mathrm{OA}}=3\)
\(\frac{4}{\mathrm{OA}}=3\)
OA = \(\frac{4}{3}\)
x-intercept = –\(\frac{4}{3}\) as point A is to the left side of Y-axis.

Question 18.
Find the distance of P(-1, 1) from the line 12(x + 6) = 5(y – 2).
Solution:
Given equation of the line is
12(x + 6) = 5(y – 2)
12x + 72 = 5y – 10
12x – 5y + 82 = 0
Let p be the perpendicular distance of the point (-1, 1) from the line 12x – 5y + 82 = 0.

Question 19.
Line through A(h, 3) and B(4,1) intersect the line lx – 9y -19 = 0 at right angle. Find the value of h.
Solution:
Given, A(h, 3) and B(4, 1)
Slope of AB (m₁) = \(\frac{1-3}{4-h}\)
m₁ = \(\frac{2}{h-4}\)
Slope of line 7x – 9y – 19 = 0 is m₂ = \(\frac{7}{9}\)
Since line AB and line 7x – 9y – 19 = 0 are perpendicular to each other,
m₁ × m₂ = -1
\(\frac{2}{h-4} \times \frac{7}{9}=-1\)
14 = 9(4 – h)
14 = 36 – 9h
9h = 22
h = \(\frac{22}{9}\)

Question 20.
Two lines passing through M(2, 3) intersect each other at an angle of 45°. If slope of one line is 2, find the equation of the other line.
Solution:
Let m be the slope of the required line which make an angle of 45° with the other line.
Slope of one of the lines is 2.
tan 45° = \(\left|\frac{\mathrm{m}-2}{1+\mathrm{m}(2)}\right|\)
1 = \(\left|\frac{m-2}{1+2 m}\right|\)
\(\frac{m-2}{1+2 m}=\pm 1\)
\(\frac{\mathrm{m}-2}{1+2 \mathrm{~m}}\) = 1 or \(\frac{\mathrm{m}-2}{1+2 \mathrm{~m}}\) = -1
m – 2 = 1 + 2m or m – 2 = -1 – 2m
m = -3 or 3m = 1
m = -3 or m = \(\frac{1}{3}\)
Required line passes through M(2, 3)
When m = -3, equation of the line is
y – 3 = -3(x – 2)
y – 3 = -3x + 6
∴ 3x + y = 9
When m = \(\frac{1}{3}\), equation of the line is
y – 3 = \(\frac{1}{3}\)(x – 2)
3y – 9 = x – 2
∴ x – 3y + 7 = 0

Question 21.
Find the y-intercept of the line whose slope is 4 and which has x-intercept 5.
Solution:
Given, slope = 4, x-intercept = 5
Since the x-intercept of the line is 5, it passes through (5, 0).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
Equation of the required line is
y – 0 = 4(x – 5)
y = 4x – 20
4x – y = 20
\(\frac{4 x}{20}-\frac{y}{20}=1\)
\(\frac{x}{5}+\frac{y}{(-20)}=1\)
This equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\), where
x-intercept = b, y-intercept = -20

Question 22.
Find the equations of the diagonals of the rectangle whose sides are contained in the lines x = 8, x = 10, y = 11 and y = 12.
Solution:
Given, equations of sides of rectangle are x = 8, x = 10, y = 11 and y = 12

From the above diagram,
Vertices of rectangle are A(8, 11), B(10, 11), C(10, 12) and D(8, 12).
Equation of diagonal AC is
\(\frac{y-11}{12-11}=\frac{x-8}{10-8}\)
\(\frac{y-11}{1}=\frac{x-8}{2}\)
2y – 22 = x – 8
x – 2y + 14 = 0
Equation of diagonal BD is
\(\frac{y-11}{12-11}=\frac{x-10}{8-10}\)
\(\frac{y-11}{1}=\frac{x-10}{-2}\)
-2y + 22 = x – 10
x + 2y = 32

Question 23.
A(1, 4), B(2, 3) and C(1, 6) are vertices of AABC. Find the equation of the altitude through B and hence find the co-ordinates of the point where this altitude cuts the side AC of ∆ABC.
Solution:
Vertices of triangle are A(1, 4), B(2, 3) and C(1, 6).
Let BD be the altitude through the vertex B.

Since both the points A and C have same x co-ordinates i.e. 1
the given points lie on a line parallel to Y-axis.
The equation of the line AC is x = 1 …..(i)
AC is parallel to Y-axis and therefore, altitude BD is parallel to X-axis.
Since the altitude BD passes through B(2, 3), the equation of altitude BD is y = 3 ……(ii)
From (i) and (ii),
Point of intersection of AC and altitude BD is (1, 3).

Question 24.
The vertices of ∆PQR are P(2, 1), Q(-2, 3) and R(4, 5). Find the equation of the median through R.
Solution:

Let S be the midpoint of side PQ.
Then RS is the median through R.
S = \(\left(\frac{2-2}{2}, \frac{3+1}{2}\right)\) = (0, 2)
The median RS passes through the points R(4, 5) and S(0, 2).
∴ Equation of median RS is
\(\frac{y-5}{2-5}=\frac{x-4}{0-4}\)
⇒ \(\frac{y-5}{-3}=\frac{x-4}{-4}\)
⇒ 4(y – 5) = 3(x – 4)
⇒ 4y – 20 = 3x – 12
∴ 3x – 4y + 8 = 0

Question 25.
A line perpendicular to segment joining A(1, 0) and B(2, 3) divides it internally in the ratio 1 : 2. Find the equation of the line. Solution:
Given, A(1, 0), B(2, 3)
Slope of AB = \(\frac{3-0}{2-1}\) = 3
Required line is perpendicular to AB.
Slope of required line = \(\frac{-1}{3}\)
Let point C divide AB in the ratio 1 : 2.

Required line passes through \(\left(\frac{4}{3}, 1\right)\) and has slope = \(\frac{-1}{3}\)
Equation of the line in slope point form is y – y₁ = m(x – x₁)
The equation of the required line is
y – 1 = \(\frac{-1}{3}\left(x-\frac{4}{3}\right)\)
⇒ 3(y – 1) = \(-1\left(x-\frac{4}{3}\right)\)
⇒ 3y – 3 = -x + \(\frac{4}{3}\)
⇒ 9y – 9 = -3x + 4
⇒ 3x + 9y = 13

Question 26.
Find the co-ordinates of the foot of the perpendicular drawn from the point P(-1, 3) to the line 3x – 4y – 16 = 0.
Solution:

Let M be the foot of perpendicular drawn from P(-1, 3) to the line 3x – 4y – 16 = 0
Slope of the line 3x – 4y – 16 = 0 is \(\frac{-3}{-4}=\frac{3}{4}\)
Since PM ⊥ to line (i),
slope of PM = \(\frac{-4}{3}\)
Equation of PM is
y – 3 = \(\frac{-4}{3}\) (x + 1)
⇒ 3(y – 3) = -4(x + 1)
⇒ 3y – 9 = -4x – 4
∴ 4x + 3y – 5 = 0 ……(ii)
The foot of perpendicular i.e., point M, is the point of intersection of equation (i) and (ii).
By (i) × 3 + (ii) × 4, we get
25x = 68
x = \(\frac{68}{25}\)
Substituting x = \(\frac{68}{25}\) in (ii), we get

The co-ordinates of the foot of perpendicular M are \(\left(\frac{68}{25}, \frac{-49}{25}\right)\)

Question 27.
Find points on the X-axis whose distance from the line \(\frac{x}{3}+\frac{y}{4}=1\) is 4 units.
Solution:
The equation of line is \(\frac{x}{3}+\frac{y}{4}=1\)
i.e. 4x + 3y – 12 = 0 …..(i)
Let (h, 0) be a point on the X-axis.
The distance of this point from line (i) is 4.
⇒ \(\frac{|4 h+3(0)-12|}{\sqrt{4^{2}+3^{2}}}=4\)
⇒ \(\frac{|4 \mathrm{~h}-12|}{5}=4\)
⇒ |4h – 12| = 20
⇒ 4h – 12 = 20 or 4h – 12 = -20
⇒ 4h = 32 or 4h = -8
⇒ h = 8 or h = -2
∴ The required points are (8, 0) and (-2, 0).

Question 28.
The perpendicular from the origin to a line meets it at (-2, 9). Find the equation of the line.
Solution:

Slope of ON = \(\frac{9-0}{-2-0}=\frac{-9}{2}\)
Since line AB ⊥ ON,
slope of the line AB perpendicular to ON is \(\frac{2}{9}\) and it passes through point N(-2, 9).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
Equation of line AB is
y – 9 = \(\frac{2}{9}\)(x + 2)
⇒ 9(y – 9) = 2(x + 2)
⇒ 9y – 81 = 2x + 4
⇒ 2x – 9y + 85 = 0

Question 29.
P(a, b) is the midpoint of a line segment intercepted between the axes. Show that the equation of the line is \(\frac{x}{a}+\frac{y}{b}=2\).
Solution:
Let the intercepts of a line AB be x₁ and y₁ on the X and Y-axes respectively.
A ≡ (x₁, 0), B = (0, y₁)
P(a, b) is the midpoint of a line segment AB intercepted between the axes.

Question 30.
Find the distance of the line 4x – y = 0 from the point P(4, 1) measured along the line making an angle of 135° with the positive X-axis.
Solution:

Let a line L make angle 135° with positive X-axis.
Required distance = PQ, where PQ || line L
Slope of PQ = tan 135°
= tan (180° – 45°)
= -tan 45°
= -1
Equation of PQ is
y – 1 = (-1)(x – 4)
y – 1 = -x + 4
x + y = 5 …..(i)
To get point Q we solve the equation 4x – y = 0 with (i)
Substituting y = 4x in (i), we get
5x = 5
x = 1
Substituting x = 1 in (i), we get
1 + y = 5
y = 4
∴ Q = (1, 4)
PQ = \(\sqrt{(4-1)^{2}+(1-4)^{2}}\)
= \(\sqrt{9+9}\)
= 3√2

Question 31.
Show that there are two lines which pass through A(3, 4) and the sum of whose intercepts is zero.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}=1\) ……(1)
This line passes through (3, 4)
\(\frac{3}{a}+\frac{4}{b}=1\) …..(ii)
Since the sum of the intercepts of the line is zero,
a + b = 0
a = -b ……(iii)
Substituting the value of a in (ii), we get
\(\frac{3}{-b}+\frac{4}{b}=1\)
\(\frac{1}{b}\) = 1
b = 1
a = -1 ……[From (iii)]
Substituting the values of a and b in (i),
the equation of the required line is
\(\frac{x}{-1}+\frac{y}{1}=1\)
x – y = -1
∴ x – y + 1 = 0

Case II: Line passing through origin.
Slope of line passing through origin and A(3, 4) is
m = \(\frac{4-0}{3-0}=\frac{4}{3}\)
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
The equation of the required line is
y = \(\frac{4}{3}\)x
∴ 4x – 3y = 0
∴ There are two lines which pass through A(3, 4) and the sum of whose intercepts is zero.

Question 32.
Show that there is only one line which passes through B(5, 5) and the sum of whose intercepts is zero.
Solution:
When line is passing through origin, the sum of intercepts made by the line is zero.
Slope of line passing through origin and B(5, 5) is
m = \(\frac{5-0}{5-0}\) = 1
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
The equation of the required line is y = x
∴ x – y = 0
∴ There is only one line which passes through B(5, 5) and the sum of whose intercepts is zero.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Trigonometry – II Ex 3.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.2

Question 1.
Find the values of:
i. sin 690°
ii. sin 495°
iii. cos 315°
iv. cos 600°
v. tan 225°
vi. tan (- 690°)
vii. sec 240°
viii. sec (- 855°)
ix. cosec 780°
x. cot (-1110°)
Solution:
i. sin 690° = sin (720° -30°)
Solution:
i. sin 690° = sin (720° -30°)
= sin (2 x 360° – 30°)
= – sin 30°
= \(\frac{-1}{2}\)

ii. sin 495° = sin (360° + 135°)
= sin (135°)
= sin (90° + 45°)
= cos 45°
= \(\frac{1}{\sqrt{2}}\)

iii. cos 315° = cos (270° + 45°)
sin 45° = \(\frac{1}{\sqrt{2}}\)

iv. cos 600° = cos (360° + 240°)
= cos 240°
= cos (180° + 60°)
= – cos 60°
= \(-\frac{1}{2}\)

v. tan 225° = tan (180° + 45°)
= tan 45°
= 1 .

vi. tan (- 690°) = – tan 690°
= – tan (720° – 30°)
= – tan (2 x 360° – 30°)
= – (- tan 30°)
= tan 30°
= \(\frac{1}{\sqrt{3}}\)

vii. sec 240° = sec (180° + 60°)
= – sec 60°
= – 2

viii. sec (-855°) = sec (855°)
= sec (720°+135°)
= sec (2 x360°+ 135°) = sec 135°
= sec (90° + 45°)
= – cosec 45°
= –\(\sqrt{2}\)

ix. cosec 780° = cosec (720° + 60°)
= cosec (2 x 360° + 60°)
= cosec 60°
= \(\frac{2}{\sqrt{3}}\)

x. cot (-1110°) =-cot (1110°)
= -cot (1080°+ 30°)
= – cot (3 x 360° + 30° )
= – cot 30°
= – \(\sqrt{3}\)

Question 2.
Prove the following:
i. \(\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x\)
ii. \(\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]\)
iii. sec 840° cot (- 945°) + sin 600° tan (- 690°) = 3/2
iv. \(\frac{{cosec}\left(90^{\circ}-x\right) \sin \left(180^{\circ}-x\right) \cot \left(360^{\circ}-x\right)}{\sec \left(180^{\circ}+x\right) \tan \left(90^{\circ}+x\right) \sin (-x)}=1\)
v. \(\frac{\sin ^{3}(\pi+x) \sec ^{2}(\pi-x) \tan (2 \pi-x)}{\cos ^{2}\left(\frac{\pi}{2}+x\right) \sin (\pi-x) {cosec}^{2}(-x)}=\tan ^{3} x\)
vi. cos θ + sin (270° + θ) – sin (270° – θ) + cos (180° + θ) = 0
Solution:
i.

ii. L.H.S.
= cos ( \(\frac{3 \pi}{2}\) + x) cos (2π + x) . [cot ( – x) + (2π + x)]
= (sin x)(cos x) (tan x + cot x)
= sin x cos x ( \(\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)\))
= sin x cos x \(\left(\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right)\)
= sin x cos x \(\left(\frac{1}{\sin x \cos x}\right)\)
= 1 = R.H.S

iii. sec 840° = sec (720° + 120°)
= sec (2 x 360° + 120°)
= sec (120°)
= sec (90° + 30°)
= – cosec 30°
= -2

cot(-945°) = -cot 945°
= -cot (720° + 225°)
= -cot (2 x 360° +225°)
= -cot (225°)
= -cot (180° + 459)
= -cot 45°
= -1

sin 600° = sin (360° + 240°)
= sin (240°)
= sin (180° +60°)
= – sin 60° = –\(\frac{\sqrt{3}}{2}\)

tan (-690°) = – tan 690°
= – tan (360° +330°)
= -tan (330°)
=- tan (360° – 30°)
=-(-tan 30°)
= tan 30°0 = \(\frac{1}{\sqrt{3}}\)

L.H.S. = sec 840° cot (-945°) + sin 600° tan (-690°)
= (-2)(-1) + \(\left(-\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{3}}\right)\)
= 2 – \(\frac{1}{2}=\frac{3}{2}\)
= R. H. S.

iv.

= 1
= R.H.S

v.

vi. L.H.S. = cos θ + sin (270° + θ) – sin (270° – θ) + cos (180° + θ)
= cos θ + (- cos θ)-(- cos θ) – cos θ
= cos θ – cos θ + cos θ – cos θ
= 0
= R.H.S.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Trigonometry – II Ex 3.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.1

Question 1.
Find the values of:
i. sin 150°
ü. cos 75°
iii. tan 105°
iv. cot 225°
Solution:
i. sin 15° = sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
\(\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)=\frac{\sqrt{3}-1}{2 \sqrt{2}}\)
[Note: Answer given in the textbook is \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\) However, as per our calculation it is \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)

ii. cos 75° = cos (45° + 30°)
= cos 45° cos 30° – sin 45° sin 30°

iii. tan 105° = tan (60° +45°)

iv. cot 225°

Question 2.
Perove the following:
i. \(\cos \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-y\right) -\sin \left(\frac{\pi}{2}-x\right) \sin \left(\frac{\pi}{2}-y\right)=-\cos (x+y)\)
Solution:
L.H.S

= -(cos x cos y – sin x sin y)
= – cos (x+y)
= R.H.S

ii. \(\tan \left(\frac{\pi}{4}+\theta\right)=\frac{1+\tan \theta}{1-\tan \theta}\)
L.H.S =\(\tan \left(\frac{\pi}{4}+\theta\right)\)

R.H.S.
[Note : The question has been modified.]

iii. \(\left(\frac{1+\tan x}{1-\tan x}\right)^{2}=\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}\)
Solution:

iv. sin [(n+1)A] . sin [(n+2)A] + cos [(n+1)A] . cos [(n+2)A] = cos A
Solution:
L.H.S. = sin [(n + 1)A] . sin [(n + 2)A] + cos [(n + 1)A] . cos [(n + 2)A]
= cos [(n + 2)A] . cos [(n + 1)A] + sin [(n + 2)A] . sin [(n + 1)A]
Let(n+2)Aaand(n+l)Ab …(i)
∴ L.H.S. = cos a. cos b + sin a. sin b
= cos (a — b)
= cos [(n + 2)A — (n + I )A]
…[From (i)]
cos[(n+2 – n – 1)A]
= cos A
= R.H.S.

v. \(\sqrt{2} \cos \left(\frac{\pi}{4}-\mathrm{A}\right)=\cos \mathrm{A}+\sin \mathrm{A}\)
Solution:

vi. \(\frac{\cos (x-y)}{\cos (x+y)}=\frac{\cot x \cot y+1}{\cot x \cot y-1}\)
Solution:

vii. cos (x + y). cos (x – y) = cos²y – sin²x
Solution:
L.H.S. = cos(x + y). cos(x – y)
= (cos x cos y – sin x sin y). (cos x cos y + sin x sin y)
= cos² x cos²y – sin²x sin²y
…[∵ (a – b) (a + b) = a² – b²]
= (1 – sin²x) cos²y – sin²x (1 – cos²y)
…[∵ sin²e + cos²0 = 1]
= cos²y – cos²y sin²x – sin²x + sin²x cos²y
= cos²y – sin²x
=R.H.S.

viii.\(\frac{\tan 5 A-\tan 3 A}{\tan 5 A+\tan 3 A}=\frac{\sin 2 A}{\sin 8 A}\)
Solution:

ix. tan 8θ – tan 5θ – tan 3θ = tan 8θ tan 5θ tan 3θ
Solution:
Since, 8θ = 5θ + 3θ
∴ tan 8θ = tan (5θ + 3θ)
∴ tan 8θ = \(\frac{\tan 5 \theta+\tan 3 \theta}{1-\tan 5 \theta \tan 3 \theta}\)
∴ tan 8θ (1 – tan 5θ.tan 3θ) = tan 5θ + tan 3θ
∴ tan 8θ – tan8θ.tan5θ.tan3θ = tan5θ + tan 3θ
∴ tan 8θ – tan 5θ – tan 3θ = tan 8θ.tan 5θ.tan 3θ

x. tan 50° = tan 40° + 2tan 10°
Solution:
Since, 50° = 10° +40°
∴ tan 50° = tan (10° + 40°)
∴ \(\frac{\tan 10^{\circ}+\tan 40^{\circ}}{1-\tan 10^{\circ} \tan 40^{\circ}}\)
∴ tan 50° (1 – tan 10° tan 40°) = tan 10° + tan 40°
∴ tan 50° – tan 10° tan 40° tan 50° = tan 10° + tan 40°
∴ tan 50° – tan 10° tan 40° tan (90° – 40°) = tan 10° + tan 40°
∴ tan 50° – tan 10° tan 40° cot 40°
= tan 10° + tan 40° …[∵ tan (90° – θ) = cot θ]
∴ tan 50° – tan 10° tan 40°. \(\frac{1}{\tan 40^{\circ}}\) = tan 10° + tan 40°
∴ tan 50° – tan 10°. 1 = tan 10° + tan 40°
∴ tan 50° = tan 40° + 2 tan 10°

xi. \(\frac{\cos 27^{\circ}+\sin 27^{\circ}}{\cos 27^{\circ}-\sin 27^{\circ}}\) = tan 72°
Solution:
\(\frac{\cos 27^{\circ}+\sin 27^{\circ}}{\cos 27^{\circ}-\sin 27^{\circ}}\)
Dividing numerator and cos 27°, we get denominator by cos 27°, we get

= tan (45° + 27°)
= tan 72° = R.H.S

xii. \(\frac{\cos 27^{\circ}+\sin 27^{\circ}}{\cos 27^{\circ}-\sin 27^{\circ}}=\tan 72^{\circ}\)
Solution:
Since 45° = 10° + 35°,
tan 45° = tan (10° +35°)
∴ \(\frac{\tan 10^{\circ}+\tan 35^{\circ}}{1-\tan 10^{\circ} \tan 35^{\circ}}\)
∴ 1 – tan 10° tan 35o = tan 10° + tan 35°
∴ tan 10° + tan 35° + tan 10° tan 35° = 1

xiii. tan 10° + tan 35° + tan 10°. tan 35° = 1
Solution:

xiv. \(\frac{\cos 15^{\circ}-\sin 15^{\circ}}{\cos 15^{\circ}+\sin 15^{\circ}}=\frac{1}{\sqrt{3}}\)
Solution:
Dividing numerator and cos 15°, we get

= tan (45° + 15°)
= tan 30° = \(\frac{1}{\sqrt{3}}\) = R.H.S

Question 3.
If sin A = \(-\frac{5}{13}\),π < A < \(\frac{3 \pi}{2}\) and cos B = \(\frac{3}{5}, \frac{3 \pi}{2}\) < B < 2π, find
i. sin (A+B)
ii. cos (A-B)
iii. tan (A + B)
Solution:
Given, sin A = \(-\frac{5}{13}\)
We know that,
cos² A = 1 – sin²A = \(1-\left(-\frac{5}{13}\right)^{2}=1-\frac{25}{169}=\frac{144}{169}\)
∴ cos A = \(\pm \frac{12}{13}\)
Since, π < A < \(\frac{3 \pi}{2}\)
∴ ‘A’ lies in the 3rd quadrant.
∴ cos A<0
cos A = \(\frac{-12}{13}\)
Also,cos B = \(\frac{3}{5}\)
∴ sin²B = 1 – cos²B = \(1-\left(\frac{3}{5}\right)^{2}=1-\frac{9}{25}=\frac{16}{25}\)
∴ sin B = \(\pm \frac{4}{5}\)
Since, \(\frac{3 \pi}{2}\) < B < 2π
∴ ‘B’ lies in the 4th quadrant.
∴ sin B<0
Sin B = \(\frac{-4}{5}\)

i. sin (A + B) = sin A cos B+cos A sin B

ii. cos (A -B) = cos A cos B + sin A sin B

iii.

Question 4.
If tan A = \(\frac{5}{6}\) , tan B = \(\frac{1}{11}\) prove that A + B = \(\frac{\pi}{4}\)
Solution:
Given tan A = \(\frac{5}{6}\), tan B = \(\frac{1}{11}\)

∴ tan (A + B) = tan \(\frac{\pi}{4}\)
∴ A + B = \(\frac{\pi}{4}\)

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 2 Trigonometry – I Miscellaneous Exercise 2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Trigonometry – I Miscellaneous Exercise 2

I. Select the correct option from the given alternatives.

Question 1.
The value of the expression
cos1°. cos2°. cos3° … cos 179° =
(A) -1
(B) 0
(C) \(\frac{1}{\sqrt{2}}\)
(D) 1
Answer:
(B) 0

Explanation:
cos 1° cos 2° cos 3° … cos 179°
= cos 1° cos 2° cos 3° … cos 90°… cos 179°
= 0 …[∵ cos 90° = 0]

Question 2.
\(\frac{\tan \mathrm{A}}{1+\sec \mathrm{A}}+\frac{1+\sec \mathrm{A}}{\tan \mathrm{A}}\) is equal to
(A) 2cosec A
(B) 2 sec A
(C) 2 sin A
(D) 2 cos A
Answer:
(A) 2cosec A

Explanation:

Question 3.
If α is a root of 25cos² θ + 5cos θ – 12 = 0, \(\frac{\pi}{2}\) < α < π, then sin 2α is equal to
(A) \(-\frac{24}{25}\)
(B) \(-\frac{13}{18}\)
(C) \(\frac{13}{18}\)
(D) \(\frac{24}{25}\)
Answer:
(A) \(-\frac{24}{25}\)

Explanation:

25 cos² θ + 5 cos θ – 12 = 0
∴ (5cos θ + 4) (5 cos θ – 3) = 0
∴ cos θ = \(-\frac{4}{5}\) or cos θ = \(\frac{3}{5}\)
Since \(\frac{\pi}{2}\) < α < π,
cos α < 0
∴ cos α = \(-\frac{4}{5}\)
sin² α = 1 – cos² α = 1 – \(\frac{16}{25}=\frac{9}{25}\)
∴ sin α = \(\pm \frac{3}{5}\)
Since \(\frac{\pi}{2}\) < α < π sin α > 0
∴ sin α = 3/5
sin 2 α = 2 sin α cos α
= \(2\left(\frac{3}{5}\right)\left(\frac{-4}{5}\right)=-\frac{24}{25}\)

Question 4.
If θ = 60°, then \(\frac{1+\tan ^{2} \theta}{2 \tan \theta}\) is equal to
(A) \(\frac{\sqrt{3}}{2}\)
(B) \(\frac{2}{\sqrt{3}}\)
(C) \(\frac{1}{\sqrt{3}}\)
(D) \(\sqrt{3}\)
Answer:
(B) \(\frac{2}{\sqrt{3}}\)

Explanation:

Question 5.
If sec θ = m and tan θ = n, then \(\frac{1}{m}\left\{(m+n)+\frac{1}{(m+n)}\right\}\) is equal to
(A) 2
(B) mn
(C) 2m
(D) 2n
Answer:
(A) 2
Explanation:

Question 6.
If cosec θ + cot θ = \(\frac{5}{2}\), then the value of tan θ is
(A) \(\frac{14}{25}\)
(B) \(\frac{20}{21}\)
(C) \(\frac{21}{20}\)
(D) \(\frac{15}{16}\)
Answer:
(B) \(\frac{20}{21}\)

Explanation:
cosec θ + cot θ = \(\frac{5}{2}\) …………….(i)
cosec² θ – cot² θ = 1
∴ (cosec θ + cot θ) (cosec θ – cot θ) = 1
∴ \(\frac{5}{2}\) (cosec θ – cot θ) = 1
∴ cosec θ – cot θ = \(\frac{2}{5}\) …(ii)
Subtracting (ii) from (i), we get
2 cot θ = \(\frac{5}{2}-\frac{2}{5}=\frac{21}{10}\)
∴ cot θ = \(\frac{21}{20}\)
∴ tan θ = \(\frac{20}{21}\)

Question 7.
\(1-\frac{\sin ^{2} \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}-\frac{\sin \theta}{1-\cos \theta}\) equals
(A) 0
(B) 1
(C) sin θ
(D) cos θ
Answer:
(D) cos θ

Explanation:

Question 8.
If cosec θ – cot θ = q, then the value of cot θ is
(A) \(\frac{2 q}{1+q^{2}}\)
(B) \(\frac{2 q}{1-q^{2}}\)
(C) \(\frac{1-\mathrm{q}^{2}}{2 \mathrm{q}}\)
(D) \(\frac{1+q^{2}}{2 q}\)
Answer:
(C) \(\frac{1-\mathrm{q}^{2}}{2 \mathrm{q}}\)

Explanation:

cosec θ – cot θ = q ……(i)
cosec² θ – cot² θ = 1
∴ (cosec θ + cot θ) (cosec θ – cot θ) = 1
∴ (cosec θ + cot θ)q = 1
∴ cosec θ + cot θ = 1/q …….(ii)
Subtracting (i) from (ii), we get
2cot θ = \(\frac{1}{\mathrm{q}}-\mathrm{q}\)
∴ cot θ = \(\frac{1-q^{2}}{2 q}\)

Question 9.
The cotangent of the angles \(\frac{\pi}{3}, \frac{\pi}{4}\) and \(\frac{\pi}{6}\) are in
(A) A.P.
(B) G.P.
(C) H.P.
(D) Not in progression
Answer:
(B) G.P.

Explanation:

Question 10.
The value of tan 1°.tan 2° tan 3° equal to
(A) -1
(B) 1
(C) \(\frac{\pi}{2}\)
(D) 2
Answer:
(B) 1

Explanation:

tan1° tan2° tan3° … tan89°
= (tan 1° tan 89°) (tan 2° tan 88°)
…(tan 44° tan 46°) tan 45°
= (tan 1 ° cot 1 °) (tan 2° cot 2°)
…(tan 44° cot 44°) . tan 45°
…tan(∵ 90° – θ) = cot θ]
= 1 x 1 x 1 x … x 1 x tan 45° =1

II. Answer the following:

Question 1.
Find the trigonometric functions of:
90°, 120°, 225°, 240°, 270°, 315°, -120°, -150°, -180°, -210°, -300°, -330°
Solution:
Angle of measure 90° :
Let m∠XOA = 90°
Its terminal arm (ray OA)
intersects the standard, unit circle at P(0, 1).

∴ x = 0 and y = 1
sin 90° = y = 1
cos 90° = x = 0
tan 90° = \(\frac{y}{x}=\frac{1}{0}\), which is not defined
cosec 90° = \(\frac{1}{y}=\frac{1}{1}\) = 1
sec 90° = \(\frac{1}{x}=\frac{1}{0}\), which is not defined
cot 90° = \(\frac{x}{y}=\frac{0}{1}\) = 0

Angle of measure 120° :
Let m∠XOA =120°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Since point P lies in the 2nd quadrant, x < 0, y > 0

[Note: Answer given in the textbook of tan 120° is \(\frac{-1}{\sqrt{3}}\) and cot 120° is \(-\sqrt{3}\). However, as per our \(-\sqrt{3}\) calculation the answer of tan 120° is \(-\sqrt{3}\) and cot 120° is \(-\frac{1}{\sqrt{3}}\)

Angle of measure 225° :
Let m∠XOA = 225°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 45° – 45° – 90° triangle.
OP = 1

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure 240° :
Let m∠XOA = 240°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure 270° :
Let m∠XOA = 270°
Its terminal arm (ray OA) intersects the standard unit circle at P(0, – 1).
x = 0 andy = – 1
sin 270° = y = -1
cos 270° = x = 0
tan 270° = \(\frac{y}{x}\)

Angle of measure 315° :
Let m∠XOA = 315°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 45° – 45° – 90° triangle.
OP = 1


[Note: Answer given in the textbook of cot 315° is 1. However, as per our calculation it is -1.]

Angle of measure (-120°):
Let m∠XOA = – 120°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1,

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure (-150°) :
Let m∠XOA = – 150°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Angle of measure (-180°):
Let m∠XOA = – 180°
Its terminal arm (ray OA) intersects the standard unit circle at P(- 1, 0).
∴ x = – 1 andy = 0
sin (-180°) = y = 0
cos (-180°) = x
= -1

Angle of measure (- 210°):
Let m∠XOA = -210°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Angle of measure (- 300°):
Let m∠XOA = – 300° Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 30° – 60° – 90° triangle.
OP = 1

Since point P lies in the 1st quadrant, x>0,y>0
x = OM = \(\frac{1}{2}\) and
y = PM = \(\frac{\sqrt{3}}{2}\)

Angle of measure (- 330°):
Let m∠XOA = – 330°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1

Since point P lies in the 1st quadrant, x > 0, y > 0
∴ x = OM = \(\frac{\sqrt{3}}{2}\) and y = PM = \(\frac{1}{2}\)

Question 2.
State the signs of:
i. cosec 520°
ii. cot 1899°
iii. sin 986°
Solution:
i. 520° =360° + 160°
∴ 520° and 160° are co-terminal angles.
Since 90° < 160° < 180°,
160° lies in the 2nd quadrant.
∴ 520° lies in the 2nd quadrant,
∴ cosec 520° is positive.

ii. 1899° = 5 x 360° + 99°
∴ 1899° and 99° are co-terminal angles.
Since 90° < 99° < 180°,
99° lies in the 2nd quadrant.
∴ 1899° lies in the 2nd quadrant.
∴ cot 1899° is negative.

iii. 986° = 2x 360° + 266°
∴ 986° and 266° are co-terminal angles.
Since 180° < 266° < 270°,
266° lies in the 3rd quadrant.
∴ 986° lies in the 3rd quadrant.
∴ sin 986° is negative.

Question 3.
State the quadrant in which 6 lies if
i. tan θ < 0 and sec θ > 0
ii. sin θ < 0 and cos θ < 0
iii. sin θ > 0 and tan θ < 0
Solution:
i. tan θ < 0 tan θ is negative in 2nd and 4th quadrants, sec θ > 0
sec θ is positive in 1st and 4th quadrants.
∴ θ lies in the 4th quadrant.

ii. sin θ < 0
sin θ is negative in 3rd and 4th quadrants, cos θ < 0
cos θ is negative in 2nd and 3rd quadrants.
.’. θ lies in the 3rd quadrant.

iii. sin θ > 0
sin θ is positive in 1st and 2nd quadrants, tan θ < 0
tan θ is negative in 2nd and 4th quadrants.
∴ θ lies in the 2nd quadrant.

Question 4.
Which is greater?
sin (1856°) or sin (2006°)
Solution:
1856° = 5 x 360° + 56°
∴ 1856° and 56° are co-terminal angles.
Since 0° < 56° < 90°, 56° lies in the 1st quadrant.
∴ 1856° lies in the 1st quadrant,
∴ sin 1856° >0 …(i)
2006° = 5 x 360° + 206°
∴ 2006° and 206° are co-terminal angles.
Since 180° < 206° < 270°,
206° lies in the 3rd quadrant.
∴ 2006° lies in the 3rd quadrant,
∴ sin 2006° <0 …(ii)
From (i) and (ii),
sin 1856° is greater.

Question 5.
Which of the following is positive?
sin(-310°) or sin(310°)
Solution:
Since 270° <310° <360°,
310° lies in the 4th quadrant.
∴ sin (310°) < 0
-310° = -360°+ 50°
∴ 50° and – 310° are co-terminal angles.
Since 0° < 50° < 90°, 50° lies in the 1st quadrant.
∴ – 310° lies in the 1st quadrant.
∴ sin (- 310°) > 0
∴ sin (- 310°) is positive.

Question 6.
Show that 1 – 2sin θ cos θ ≥ 0 for all θ ∈ R.
Solution:
1 – 2 sin θ cos θ
= sin² θ + cos² θ – 2sin θ cos θ
= (sin θ – cos θ)² ≥ 0 for all θ ∈ R

Question 7.
Show that tan² θ + cot² θ ≥ 2 for all θ ∈ R.
Solution:

Question 8.
If sin θ = \(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\), then find the values of cos θ, tan θ in terms of x and y.
Solution:
Given, sin θ = \(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\)
we know that
cos²θ = 1 – sin² θ

[Note: Answer given in the textbook of cos θ = \(\frac{2 x y}{x^{2}+y^{2}}\) and tan θ = \(. However, as per our calculation the answer of cos θ = ± [latex]\frac{2 x y}{x^{2}+y^{2}}\) and tan θ = ± \(\frac{x^{2}-y^{2}}{2 x y}\). ]

Question 9.
If sec θ = \(\sqrt{2}\) and \(\frac{3 \pi}{2}\) < θ < 2π, then evaluate \(\frac{1+\tan \theta+{cosec} \theta}{1+\cot \theta-{cosec} \theta}\)
Solution:
Given sec θ = \(\sqrt{2}\)
We know that,
tan² θ = sec² θ – 1
= (\(\sqrt{2}\)) – 1
= 2 – 1 = 1
∴ tan θ = ±1
Since \(\frac{3 \pi}{2}\) < θ < 2π
θ lies in the 4th quadrant.
∴ tan θ < 0
∴ tan θ = -1

Question 10.
Prove the following:

i. sin²A cos² B + cos²A sin²B + cos²A cos²B + sin²A sin²B = 1
Solution:
L.H.S. = sin²A cos² B + cos²A sin²B + cos²A cos²B + sin²A sin²B
= sin²A (cos² B + sin² B) + cos² A (sin² B + cos² B)
= sin²A(1) + cos²A(1)
= 1 = R.H.S.

ii. \(\frac{(1+\cot \theta+\tan \theta)(\sin \theta-\cos \theta)}{\sec ^{3} \theta-{cosec}^{3} \theta}=\sin ^{2} \theta \cos ^{2} \theta\)
Solution:

iii. L.H.S. = \(\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}+\left(\tan \theta-\frac{1}{\cos \theta}\right)^{2}=2\left(\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta}\right)\)
Solution:
L.H.S. = \(\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}+\left(\tan \theta-\frac{1}{\cos \theta}\right)^{2}\)
= (tanθ + secθ)² + (tanθ – secθ)²
= tan² θ + 2 tan θ sec θ + sec² θ
+ tan² θ – 2 tan θ sec θ +.sec² θ
= 2(tan² θ + sec² θ)

iv. 2.sec² θ – sec⁴ θ – 2.cosec² θ + cosec⁴ θ = cot⁴ θ – tan⁴ θ
Solution:
LHS.
= 2.sec² θ – sec⁴ θ – 2.cosec² θ + cosec⁴ θ =  = 2 sec² θ – (sec² θ)² – 2cosec² θ + (cosec² θ)²
= 2(1+ tan² θ) – (1+ tan² θ)² – 2(1+ cot² θ)
+ (1+ cot² θ)²
= 2 + 2tan² θ – (1 + 2tan² θ + tan⁴ θ)
– 2 – 2cot² θ + 1 + 2cot² θ + cot⁴ θ
= 2 + 2.tan² θ – 1 – 2 tan² θ – tan⁴ θ – 2
– 2 cot² θ + 1 + 2 cot² θ + cot⁴ θ
= cot⁴ θ – tan⁴ θ = R.H.S.

v. sin⁴ θ + cos⁴ θ = sin⁴ θ + cos⁴ θ
Solution:
L.H.S. = sin⁴ θ + cos⁴ θ
= (sin² θ)² + (cos² θ)² = (sin² θ + cos² θ)² – 2sin² θ cos² θ
… [ v a² + b² = (a + b)² – 2ab]
= 1 – 2sin² θ cos² θ
= R.H.S.

vi. 2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1 = 0
L.H.S =
2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1=0
= sin⁶ θ + cos⁶ θ
= (sin² θ)³ + (cos² θ)³ = (sin² θ + cos² θ)³
– 3 sin² θ cos² θ (sin2 0 + cos2 0)
…[••• a³ + b³ = (a + b)³ – 3ab(a + b)]
= (1)³ – 3 sin² θ cos² θ(1)
= 1-3 sin² θ cos² θ sin⁴ θ + cos⁴ θ
= (sin² θ)² + (cos² θ)² = (sin² θ + cos² θ)² – 2 sin² θ cos² θ
…[Y a² + b² = (a + b)² – 2ab]
= 1-2 sin² θ cos² θ
L.H.S.= 2(sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ) + 1
= 2(1-3 sin² θ cos² θ) -3(1 – 2 sin² θ cos² θ) + 1
= 2-6 sin² θ cos² θ – 3 + 6 sin² θ cos² θ + 1 = c
= R.H.S.

vii. cos⁴ θ – sin⁴ θ + 1 = 2cos²θ
L.H.S. = cos⁴ θ – sin⁴ θ + 1
= (cos² θ)² – (sin² θ)² + 1 = (cos²θ + sin²θ) c(os² θ – sin²θ) +1
= (1) (cos²θ – sin²θ) + 1 = cos² θ + (1 – sin²θ)
= cos² θ + cos²θ = 2cos²θ = R.H.S.

viii. sin⁴θ + 2sin²θ cos²θ = 1 – cos⁴θ
L.H.S. = sin⁴θ + 2sin²θ cos²θ = sin²θ(sin²θ + 2cos²θ)
= (sin²θ) (sin²θ + cos²θ + cos²θ) = (1 – cos²θ) (1 + cos²θ)
= 1 – cos⁴θ = R.H.S.

ix. \(\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta+\cos \theta}+\frac{\sin ^{3} \theta-\cos ^{3} \theta}{\sin \theta-\cos \theta}=2\)
Solution:

= (sin² θ + cos² θ – sin θ cos θ) + (sin² θ + cos² θ + sinθ cosθ)
= 2 (sin² θ + cos² θ)
= 2(1)
= 2 = R.H.S.

x. tan² θ – sin² θ = sin⁴ θ sec² θ
Solution:
L.H.S. = tan² θ – sin² θ
= \(\frac{\sin ^{2} \theta}{\cos ^{2} \theta}\) – sin²θ
= sin² θ (\(\frac{1}{\cos ^{2} \theta}-1 \))
= \(\frac{\sin ^{2} \theta\left(1-\cos ^{2} \theta\right)}{\cos ^{2} \theta}\)
= (sin² θ) (sin² θ)sec² θ
= sin⁴ θ sec² θ
= R.H.S

xi. (sinθ + cosecθ)² + (cos θ + see θ)² = tan² θ + cot² θ + 7
Solution:
L.H.S. = (sinθ + cosecθ)² + (cos θ + see θ)²
= sin ² θ + cosec² θ + 2sinθ cosec θ
+ cos² θ + sec² θ + 2sec0 cos0
= (sin² θ + cos² θ) + cosec² θ + 2 + sec² θ + 2
= 1 + (1 + cot² θ) + 2 + (1 + tan² θ) + 2 = tan² θ + cot² θ + 7
= R.H.S.

xii. sin⁸θ – cos⁸θ = (sin² θ – cos² θ) (1 – 2sin² θ cos² θ)
Solution:
L.H.S. = sin⁸θ – cos⁸θ
= (sin⁴θ)² – (cos⁴θ)²
= (sin⁴θ – cos⁴θ) (sin⁴θ + cos⁴θ)
= [(sin² θ)² – (cos² θ)² ]
. [(sin² θ)² + (cos² θ)² ]
= (sin² θ + cos² θ) (sin² θ – cos² θ). [(sin² θ + cos² θ)² – 2sin² θ.cos² θ] …[Y a² + b² = (a + b)² – 2ab]
= (1) (sin² θ – cos² θ) (1² – 2sin² θ cos² θ)
= (sin² θ – cos² θ) (1 – 2sin² θ cos² θ)
= R.H.S.

xiii. sin⁶A + cos⁶A = 1 – 3 sin²A + 3sin⁴A
Soluiton:
L.H.S. = sin⁶A + cos⁶A
= (sin² A)³ + (cos² A)³
= (sin² A + cos² A)³
– 3sin²A cos²A(sin² A + cos² A)
…[ a³ + b³ = (a + b)³ – 3ab(a + b)]
= 1³ – 3sin²A cos²A (1)
= 1 – 3sin²A cos²A
= 1 – 3 sin²A (1 – sin²A)
= 1 – 3 sin²A + 3sin⁴A
= R.H.S.

xiv. (1 + tanA tanB)² + (tanA – tanB)² = sec ²A sec²B
Solution:
L.H.S. = (1 + tanA tanB)² + (tanA – tanB)²
= 1 + 2tanA tanB + tan²A tan² + tan² A- 2tanA tanB + tan²B
= 1 + tan²A + tan² B + tan²A tan²B
= 1(1+ tan²A) + tan² B(1 + tan²A)
= (1 + tan²A) (1 + tan²B)
= sec²A sec²B = R.H.S.

xv. \(\frac{1+\cot \theta+{cosec} \theta}{1-\cot \theta+{cosec} \theta}=\frac{{cosec} \theta+\cot \theta-1}{\cot \theta-{cosec} \theta+1}\)
Solution:
We know that cosec²θ – cot² θ = 1
∴ (cosec θ – cot θ) (cosec θ + cot θ) = 1

xvi. \(\frac{\tan \theta+\sec \theta-1}{\tan \theta+\sec \theta+1}=\frac{\tan \theta}{\sec \theta+1}\)
Solution:
We know that
tan²θ = sec² θ – 1
∴ tan θ. tanθ = (sec θ + 1)(sec θ – 1)

xvii. \(\frac{{cosec} \theta+\cot \theta-1}{{cosec} \theta+\cot \theta+1}=\frac{1-\sin \theta}{\cos \theta}\)
Solution:
We know that,
cot² θ = cosec² θ – 1
∴ cot θ . cot θ = (cosec θ + 1)(cosec θ – 1)

Alternate Method:

xviii. \(\frac{{cosec} \theta+\cot \theta+1}{\cot \theta+{cosec} \theta-1}=\frac{\cot \theta}{{cosec} \theta-1}\)
solution:
We know that,
cot² θ = cosec² θ – 1
∴ cot θ.cot θ = (cosec θ + 1) (cosec θ – 1)

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 2 Trigonometry – I Ex 2.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Trigonometry – I Ex 2.2

Question 1.
If 2sin A = 1 = \(\sqrt{2}\) cos B and \(\frac{\pi}{2}\) < A < π, \(\frac{3 \pi}{2}\)
Solution:
Given, 2sin A = 1
∴ sin A = 1/2
we know that,
cos² A = 1 – sin² A = 1 – \(\left(\frac{1}{2}\right)^{2}=1-\frac{1}{4}=\frac{3}{4}\)
∴ cos A = \(\pm \frac{\sqrt{3}}{2}\)
Since \(\frac{\pi}{2}\) < A < π
A lies in the 2nd quadrant.

We know that,
Sin² B = 1 – cos² B = 1 – \(\left(\frac{1}{\sqrt{2}}\right)^{2}\)\(\frac{1}{2}=\frac{1}{2}\)
∴ sin B = \(\pm \frac{1}{\sqrt{2}}\)
Since \(\frac{3 \pi}{2}\) < B < 2π
B lies in the 4th quadrant,

Question 2.
If \(\) and A, B are angles in the second quadran, then prove that 4cosA + 3 cos B = -5
Solution:
Given, \(\frac{\sin \mathrm{A}}{3}=\frac{\sin \mathrm{B}}{4}=\frac{1}{5}\)
∴ sin A = \(\frac{3}{5}\) and sin B = \(\frac{4}{5}\)
We know that,
cos² A = 1 – sin² = 1 – \(\left(\frac{3}{5}\right)^{2}\) = 1 – \(\frac{9}{25}=\frac{16}{25}\)
∴ Cos A = ± \([{4}{5}\)
Since A lies in the second quadrant,
cos A < 0
∴ Cos A = –\(\frac{4}{5}\)
Sin B = 4/5
We know that,
cos²B = 1 – sin²B = 1 – \(\left(\frac{4}{5}\right)^{2}=1-\frac{16}{25}=\frac{9}{25}\)
∴ Cos B = ±\(\frac{4}{5}\)
Since B lies in the second quadrant, cos B < 0

Question 3.
If tan θ = \(\frac{1}{2}\), evaluate \(\frac{2 \sin \theta+3 \cos \theta}{4 \cos \theta+3 \sin \theta}\)
Solution:

Question 4.
Eliminate 0 from the following:
i. x = 3sec θ, y = 4tan θ
ii. x = 6cosec θ,y = 8cot θ
iii. x = 4cos θ – 5sin θ, y = 4sin θ + 5cos θ
iv. x = 5 + 6 cosec θ,y = 3 + 8 cot θ
v. x = 3 – 4tan θ,3y = 5 + 3sec θ
Solution:
i. x = 3sec θ, y = 4tan θ
∴ sec θ = \(\frac{x}{3}\) and tan θ= \(\frac{y}{4}\)
We know that,
sec²θ – tan²θ = 1

∴ 16x² – 9y² = 144

ii. x = 6cosec θ and y = 8cot θ
.’. cosec θ = \(\) and cot θ = \(\)
We know that,
cosec² θ – cot² θ =

16x² – 9y² = 576

iii. x = 4cos θ – 5 sin θ … (i)
y = 4sin θ + 5cos θ .. .(ii)
Squaring (i) and (ii) and adding, we get
x² + y² = (4cos θ – 5sin θ)² + (4sin θ + 5cos θ)²
= 16cos²θ – 40 sinθ cosθ + 25 sin²θ + 16 sin² θ + 40sin θ cos θ + 25 cos² θ
= 16(sin² θ + cos² θ) + 25(sin² θ + cos² θ)
= 16(1) + 25(1)
= 41

iv. x = 5 + 6cosec θ andy = 3 + 8cot θ
∴ x – 5 = 6cosec θ and y – 3 = 8cot θ
∴ cosec θ = \(\frac{x-5}{6}\) and cot θ = \(\frac{y-3}{8}\)
We know that,
cosec² θ – cot² θ = 1
∴ \(\left(\frac{x-5}{6}\right)^{2}-\left(\frac{y-3}{8}\right)^{2}\) = 1

v. 2x = 3 – 4tan θ and 3y = 5 + 3sec θ
∴ 2x – 3 = -4tan θ and 3y – 5 = 3sec θ
∴ tan θ = \(\frac{3-2 x}{4}\) and sec θ = \(\frac{3 y-5}{3}\)θ
We know that, sec² θ – tan² θ = 1
∴ \(\left(\frac{3 y-5}{3}\right)^{2}-\left(\frac{3-2 x}{4}\right)^{2}\) = 1
∴ \(\left(\frac{3 y-5}{3}\right)^{2}-\left(\frac{2 x-3}{4}\right)^{2}\) = 1

Question 5.
If 2sin² θ + 3sin θ = 0, find the permissible values of cosθ.
Solution:
2sin² θ + 3sin θ = 0
∴ sin θ (2sin θ + 3) = 0
∴ sin θ = 0 or sin θ = \(\frac{-3}{2}\)
Since – 1 ≤ sin θ ≤ 1,
sin θ = 0
\(\sqrt{1-\cos ^{2} \theta}\) = 0 …[ ∵ sin² θ = 1- cos² θ]
∴ 1 – cos² θ = 0
∴ cos² θ = 1
∴ cos θ = ±1 …[∵ – 1 ≤ cos θ ≤ 1]

Question 6.
If 2cos² θ – 11 cos θ + 5 = 0, then find the possible values of cos θ.
Solution:
2cos²θ – 11 cos θ + 5 = 0
∴ 2cos² θ – 10 cos θ – cos θ + 5 = 0
∴ 2cos θ(cos θ – 5) – 1 (cos θ – 5) = 0
∴ (cos θ – 5) (2cos θ – 1) = 0
cos θ – 5 = 0 or 2cos θ – 1 = 0
∴ cos θ = 5 or cos θ = 1/2
Since, -1 ≤ cos θ ≤ 1
∴ cos θ = 1/2

Question 7.
Find the acute angle θ such 2cos² θ = 3sin θ.
Solution:
2cos²0 = 3sin θ
∴ 2(1 – sin² θ) = 3sin θ
∴ 2 – 2sin² θ = 3sin θ
∴ 2sin² θ + 3sin 9-2 = θ
∴ 2sin² θ + 4sin θ – sin θ – 2 = θ
∴ 2sin θ(sin θ + 2) -1 (sin θ + 2) = θ
∴ (sin θ + 2) (2sin θ – 1) = 0
∴ sin θ + 2 = 0 or 2sin θ – 1 = 0
∴ sin θ = -2 or sin θ = 1/2
Since, -1 ≤ sin θ ≤ 1
∴ Sin θ = 1/2
∴ θ = 30° …[ ∵ sin 30 = 1/2]

Question 8.
Find the acute angle 0 such that 5tan2 0 + 3 = 9sec 0.
Solution:
5tan² θ + 3 = 9sec θ
∴ 5(sec² θ – 1) + 3 = 9sec θ
∴ 5sec² θ – 5 + 3 = 9sec θ
∴ 5sec² θ – 9sec θ – 2 = 0
∴ 5sec² θ – 10 sec θ + sec θ – 2 = 0
∴ 5sec θ(sec θ – 2) + 1(sec θ – 2) = 0
∴ (sec θ – 2) (5sec θ + 1) = 0
∴ sec θ – 2 = 0 or 5sec θ + 1 = 0
∴ sec θ = 2 or sec θ = -1/5
Since sec θ ≥ 1 or sec θ ≤ -1,
sec θ = 2
∴ θ = 60° … [ ∵ sec 60° = 2]

Question 9.
Find sin θ such that 3cos θ + 4sin θ = 4.
Solution:
3cos θ + 4sin θ = 4
∴ 3cos θ = 4(1 – sin θ)
Squaring both the sides, we get .
9cos²θ = 16(1 – sin θ)²
∴ 9(1 – sin² θ) = 16(1 + sin² θ – 2sin θ)
∴ 9 – 9sin² θ = 16 + 16sin² θ – 32sin θ
∴ 25sin² θ – 32sin θ + 7 = 0
∴ 25sin² θ – 25sin θ – 7sin θ + 7 = 0
25sin θ (sin θ – 1) – 7 (sin θ – 1) = 0
∴ (sin θ – 1) (25sin θ – 7) = 0
∴ sin θ – 1 = 0 or 25 sin θ – 7 = 0
∴ sin θ = 1 or sin θ = \(\frac{7}{25}\)
Since, -1 ≤ sin θ ≤ 1
∴ sin θ = 1 or \(\frac{7}{25}\)
[Note: Answer given in the textbook is 1. However, as per our calculation it is 1 or \(\frac{7}{25}\).]

Question 10.
If cosec θ + cot θ = 5, then evaluate sec θ.
Solution:
cosec θ + cot θ = 5
∴ \(\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=5\)
∴ \(\frac{1+\cos \theta}{\sin \theta}=5\)
∴ 1 + cos θ = 5.sin θ
Squaring both the sides, we get
1 + 2 cos θ + cos² θ = 25 sin² θ
∴ cos² θ + 2 cos θ + 1 = 25 (1 – cos² θ)
∴ cos² θ + 2 cos θ + 1 = 25 – 25 cos² θ
∴ 26 cos² θ + 2 cos θ – 24 = 0
∴ 26 cos² θ + 26 cos θ – 24 cos θ – 24 = 0
∴ 26 cos θ (cos θ + 1) – 24 (cos θ + 1) = 0
∴ (cos θ + 1) (26 cos θ – 24) = 0
∴ cos θ + 1 = θ or 26 cos θ – 24 = 0
∴ cos θ = -1 or cos θ = \(\frac{24}{26}=\frac{12}{13}\)
When cos θ = -1, sin θ = 0
∴ cot θ and cosec x are not defined,
∴ cos θ ≠ -1
∴ cos θ = \(\frac{12}{13}\)
∴ sec θ = \(\frac{1}{\cos \theta}=\frac{13}{12}\)
[Note: Answer given in the textbook is -1 or \(\frac{13}{12}\).
However, as per our calculation it is only \(\frac{13}{12}\).]

Question 11.
If cot θ = \(\frac{3}{4}\) and π < θ < \(\frac{3 \pi}{2}\), then find the value of 4 cosec θ + 5 cos θ.
Solution:
We know that,
cosec²θ = 1 + cot² θ = \(\left(\frac{3}{4}\right)^{2}\) = 1 + \(\frac{9}{16}\)
∴ cosec² θ = \(\frac{25}{16}\)
∴ cosec θ = \(\pm \frac{5}{4}\)
Since π < θ < \(\frac{3 \pi}{2}\)
θ lies in the third quadrant.
∴ cosec θ < 0
∴ cosec θ = –\(\frac{5}{4}\)
cot θ = \(\frac{3}{4}\)
tan θ = \(\frac{1}{\cot \theta}=\frac{4}{3}\)
We know that,
sec² θ = 1 + tan² θ = 1 + \(\left(\frac{4}{3}\right)^{2}\)
= 1 + \(\frac{16}{9}=\frac{25}{9}\)
∴ sec θ = ±\(\frac{5}{3}\)
Since θ lies in the third quadrant,
sec θ < 0
∴ sec θ = –\(\frac{5}{3}\)
cos θ = \(\frac{1}{\sec \theta}=\frac{-3}{5}\)
∴ 4cosec θ + 5cos θ
= \(4\left(-\frac{5}{4}\right)+5\left(-\frac{3}{5}\right)\)
= -5 – 3 = -8
[Note: The question has been modified.]

Question 12.
Find the Cartesian co-ordinates of points whose polar co-ordinates are:
i. (3, 90°) ii. (1, 180°)
Solution:
i. (r, θ) = (3, 90°)
Using x = r cos θ and y = r sin θ, where (x, y) are the required cartesian co-ordinates, we get
x = 3cos 90° and y = 3sin 90°
∴ x = 3(0) = 0 and y = 3(1) = 3
∴ the required cartesian co-ordinates are (0, 3).

ii. (r, θ) = (1, 180°)
Using x = r cos θ and y = r sin θ, where (x, y) are the required cartesian co-ordinates, we get
x = 1(cos 180°) and y = 1(sin 180°)
∴ x = -1 and y = 0
∴ the required cartesian co-ordinates are (-1, 0).

Question 13.
Find the polar co-ordinates of points whose Cartesian co-ordinates are:
1. (5, 5) ii. (1, \(\sqrt{3}\))
ii. (-1, -1) iv. (-\(\sqrt{3}\), 1)
Solution:
i. (x, y) = (5, 5)
∴ r = \(\sqrt{x^{2}+y^{2}}\) = \(\sqrt{25+25}\)
\(=\sqrt{50}=5 \sqrt{2}\)
tan θ = \(\frac{y}{x}=\frac{5}{5}\) = 1
Since the given point lies in the 1st quadrant,
θ = 45° …[∵ tan 45° = 1]
∴ the required polar co-ordinates are (\(5 \sqrt{2}\), 45°).

ii. (x, y) = ( 1, \(\sqrt{3}\))
∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{1+3}=\sqrt{4}=2\)
tan θ = \(\frac{y}{x}=\frac{\sqrt{3}}{1}=\sqrt{3}\)
Since the given point lies in the 1st quadrant,
θ = 60° …[∵ tan 60° = \(\sqrt{3}\)]
∴ the required polar co-ordinates are (2, 60°).

iii. (x, y) = (-1, -1)
∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{1+1}=\sqrt{2}\)
tan θ = \(\frac{y}{x}=\frac{-1}{-1}=1\)
∴ tan θ = tan \(\frac{\pi}{4}\)
Since the given point lies in the 3rd quadrant,
tan θ = tan \(\left(\pi+\frac{\pi}{4}\right)\) …[∵ tan (n + x) = tanx]
∴ tan θ = tan \(\left(\frac{5 \pi}{4}\right)\)
∴ θ = \(\frac{5 \pi}{4}\) = 225°
∴ the required polar co-ordinates are (\(\sqrt{2}\), 225°).

iv. (x, y) = (-\(\sqrt{3}\) , 1)
∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{3+1}=\sqrt{4}=2\)
tan θ = \(\frac{y}{x}=\frac{1}{-\sqrt{3}}\) = -tan \(\frac{\pi}{6}\)
Since the given point lies in the 2nd quadrant,
tan θ = tan \(\left(\pi-\frac{\pi}{6}\right)\) …[∵ tan (π – x) = – tanx]
∴ tan θ = tan \(\left(\frac{5 \pi}{6}\right)\)
∴ θ = \(\frac{5 \pi}{6}\) = 150°
∴ the required polar co-ordinates are (2, 150°)

Question 14.
Find the values of:
i. sin\(\frac{19 \pi^{e}}{3}\)
ii. cos 1140°
iii. cot \(\frac{25 \pi^{e}}{3}\)
Solution:
i. We know that sine function is periodic with period 2π.
sin \(\frac{19 \pi}{3}\) = sin \(\left(6 \pi+\frac{\pi}{3}\right)\) = sin \(\frac{\pi}{3}=\frac{\sqrt{3}}{2}\)

ii. We know that cosine function is periodic with period 2π.
cos 1140° = cos (3 × 360° + 60°)
= cos 60° = \(\frac {1}{2}\)

iii. We know that cotangent function is periodic with period π.
cot \(\frac{25 \pi}{3}\) = cot \(\left(8 \pi+\frac{\pi}{3}\right)\) = cot \(\frac{\pi}{3}\) = \(\frac{1}{\sqrt{3}}\)
dhana work.txt
Displaying dhana work.txt.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 2 Trigonometry – I Ex 2.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Trigonometry – I Ex 2.1

Question 1.
Find the trigonometric functions of 0°, 30°, 45°, 60°, 150°, 180°, 210°, 300°, 330°, – 30°, – 45°, – 60°, – 90°, – 120°, – 225°, – 240°, – 270°, – 315°
Solution:
Angle of measure 0°:

Let m∠XOA = 0° = 0^c
Its terminal arm (ray OA) intersects the standard
unit circle in P(1, 0).
Hence,x = 1 and y = 0
sin 0° = y = 0,
cos 0° = x = 1,
tan 0° = \(\frac{y}{x}=\frac{0}{1}\) = 0
cot 0° = \(\frac{x}{y}=\frac{1}{0}\) which is not defined
sec 0° = \(\frac{1}{x}=\frac{1}{1}\) = 1
cot 0° = \(\frac{1}{y}=\frac{1}{0}\) which is not defined,

Angle of measure 30°:
Let m∠XOA = 30°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y)
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1

Since point P lies in 1st quadrant, x > 0, y > 0
∴ x = OM = \(\frac{\sqrt{3}}{2}\) and y = PM = \(\frac{1}{2}\)

Angle of measure 45°:
Let m∠XOA = 45°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 45° – 45° – 90° triangle.
OP = 1,

Since point P lies in the 1st quadrant, x > 0, y > 0
∴ x = OM = \(\frac{1}{\sqrt{2}}\) and
y = PM = \(\frac{1}{\sqrt{2}}\)
∴ P = (\(\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\))

Angle of measure 60°:
Let m∠XOA = 60°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1,

Angle of measure 150°:
Let m∠XOA = 150°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1,

Since point P lies in the 2nd quadrant, x < 0, y > 0

Angle of measure 180°:
Let m∠XOA = 180°
Its terminal arm (ray OA) intersects the standard unit circle at P(-1, 0).
∴ x = – 1 and y = 0
sin 180° =y = 0
cos 180° = x = -1

tan 180° = \(\frac{y}{x}\)
= \(\frac{0}{-1}\) = 0
Cosec 180° = \(\frac{1}{y}\)
= \(\frac{1}{0}\)
which is not defined.
sec 180°= \(\frac{1}{x}=\frac{1}{-1}\) = -1
cot 180° = \(\frac{x}{y}=\frac{-1}{0}\) , which is not defined.

Angle of measure 210°:
Let m∠XOA = 210°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1,

Since point P lies in the 3rd quadrant, x < 0,y < 0
∴ x = -OM = \(\frac{-\sqrt{3}}{2}\) and y = -PM = \(\frac{-1}{2}\)
∴ P ≡( \(\frac{-\sqrt{3}}{2}, \frac{-1}{2}\) )

Angle of measure 300°:
Let m∠XOA = 300°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1,

Since point P lies in the 1st quadrant, x > 0,y > 0
x = OM = \(\frac{1}{2}\) = and y = -PM = \(\frac{-\sqrt{3}}{2}\)
sin 300° = y = \(\frac{-\sqrt{3}}{2}\)
cos 300° = x = \(\frac{1}{2}\)
tan 300° = \(\frac{y}{x}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}\)

Angle of measure 330°:
Let m∠XOA = 330°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1,

Since point P lies in the 4th quadrant, x > 0, y < 0

Angle of measure 30°
Let m∠XOA = -30°
Its terminal arm (ray OA) intersects the standard unit circle at P(x,y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60 — 90° triangle.
op = 1,

Since point P lies in the 4th quadrant x > 0, y < 0

Angle of measure 45°:
Let m∠XOA = 45°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 45° – 45° – 90° triangle.
OP = 1,

Since point P lies in the 4th quadrant x > 0, y < 0

[Note : Answer given in the textbook of sin (45°) = – 1/2. However, as per our calculation it is \(-\frac{1}{\sqrt{2}}\) ]

Angle of measure (-60°):
Let m∠XOA = -60°
Its terminal arm (ray OA) intersects the standard
unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 30° – 60° – 90° triangle.
OP = 1,

Since point P lies in the 4’ quadrant,
x > 0, y < 0
x = OM =\(\frac{1}{2}\) and y = -PM = \(-\frac{\sqrt{3}}{2}\)

Angle of measure (-90°):
Let m∠XOA = -90°
It terminal arm (ray OA) intersects the standard unit circle at P(0, -1)
∴ x = 0 and y = -1
sin (-90°) = y = -1
cos (-90°) = s = 0

Angle of measure (-120°):
Let m∠XOA = – 120°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1,

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure (- 225°):
Let m∠XOA = – 225°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 45° – 45° – 90° triangle.
OP = 1,

Since point P lies in the 2nd quadrant, x < 0, y > 0

Angle of measure 2400):
Let m∠XOA = 240°
Its terminal arm (ray OA) intersects the standard
unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30°  – 60° –  900 triangle.

Since point P lies in the 2nd quadrant, x<0, y>0

Angle of measure (- 270°):
Let m∠XOA = – 270°
Its terminal arm (ray OA)
intersects the standard unit,
circle at P(0, 1).
∴ x = 0 and y = 1
sin (- 270°) = y = 1
cos (- 270°) = x = 0
tan(-270°)= \(\frac{y}{x}=\frac{1}{0}\)
which is not defined.

Angle of measure ( 315°):
Let m∠XOA 315°
Its terminal arm (ray OA) intersects the standard unit circle at P(x,y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 45° – 45° – 90° triangle.
OP = 1,

Question 2.
State the signs of:
i. tan 380°
ii. cot 230°
iii 468°
Solution:
1. 380° = 360° + 20°
∴ 380° and 20° are co-terminal angles.
Since 0° < 20° <90°0,
20° lies in the l quadrant.
∴ 380° lies in the 1st quadrant,
∴ tan 380° is positive.

ii. Since, 180° <230° <270°
∴ 230° lies in the 3rd quadrant.
∴ cot 230° is positive.

iii. 468° = 360°+108°
∴ 468° and 108° are co-terminal angles.
Since 90° < 108° < 180°,
108° lies in the 2nd quadrant.
∴ 468° lies in the 2nd quadrant.
∴ sec 468° is negative.

Question 3.
State the signs of cos 4^c and cos 4°. Which of these two functions is greater?
Solution:
Since 0° < 4° < 90°, 4° lies in the first quadrant. ∴ cos4° >0 …(i)
Since 1^c = 57° nearly,
180° < 4^c < 270°
∴ 4^c lies in the third quadrant.
∴ cos 4^c < 0 ………(ii)
From (i) and (ii),
cos 4° is greater.

Question 4.
State the quadrant in which 6 lies if
i. sin θ < 0 and tan θ > 0
ii. cos θ < 0 and tan θ > 0
Solution:
i. sin θ < 0 sin θ is negative in 3rd and 4th quadrants, tan 0 > 0
tan θ is positive in 1st and 3rd quadrants.
∴ θ lies in the 3rd quadrant.

ii. cos θ < 0 cos θ is negative in 2nd and 3rd quadrants, tan 0 > 0
tan θ is positive in 1st and 3rd quadrants.
∴ θ lies in the 3rd quadrant.

Question 5.
Evaluate each of the following:
i. sin 30° + cos 45° + tan 180°
ii. cosec 45° + cot 45° + tan 0°
iii. sin 30° x cos 45° x lies tan 360°
Solution:
i. We know that,
sin30° = 1/2, cos 45° = \(\frac{1}{\sqrt{2}}\) =, tan 180° = 0
sin30° + cos 45° +tan 180°
= \(\frac{1}{2}+\frac{1}{\sqrt{2}}+0=\frac{\sqrt{2}+1}{2}\)

ii. We know that,
cosec 45° = \(\sqrt{2}\) , cot 45° = 1, tan 0° = 0
cosec 45° + cot 45° + tan 0°
= \(\sqrt{2}\) + 1 + 0 = \(\sqrt{2}\) + 1

iii. We know that,
sin 30° = \(\frac{1}{2}\), cos 45° = \(\frac{1}{\sqrt{2}}\) =, tan 360° = 0
sin 30° x cos 45° x tan 360°
= \(\left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\right)\) = 0

Question 6.
Find all trigonometric functions of angle in standard position whose terminal arm passes through point (3, – 4).
Solution:
Let θ be the measure of the angle in standard position whose terminal arm passes through P(3, -4).
∴ x = 3 and y = -4
r = OP

Question 7.
If cos θ = \(\frac{12}{13}, 0<\theta<\frac{\pi}{2}\) find the value of \(\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta}, \frac{1}{\tan ^{2} \theta}\)
Solution:
cos θ = \(\frac{12}{13}\)
We know that,
sin² θ = 1 – cos²θ

∴ sin θ = ± \(\frac{5}{13}\)
Since 0 < θ < \(\frac{\pi}{2}\) , θ lies in the 1st quadrant, ∴ sin θ > 0

Question 8.
Using tables evaluate the following:
i. 4 cot 45° – sec2 60° + sin 30°
ii.\(\cos ^{2} 0+\cos ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}+\cos ^{2} \frac{\pi}{2}\)
Solution:
i. We know that,
cot 45° = 1, sec 60° = 2, sin 30° = 1/2
4 cot 45° – sec² 60° + sin 30°
= 4(1) – (2)² + \(\frac{1}{2}\)
= 4 – 4 + \(\frac{1}{2}=\frac{1}{2}\)

ii. We know that,

Question 9.
Find the other trigonometric functions if
i. cot θ = \(-\frac{3}{5}\), and 180 < θ < 270
ii. Sec A = \(-\frac{25}{7}\) and A lies in the second quadrant.
iii cot x = \(\frac{3}{4}\), x lies in the third quadrant.
iv. tan x = \(\frac{-5}{12}\) x lies in the fourth quadrant.
Solution:
i. cot θ = \(-\frac{3}{5}\)
we know that,
sin²θ = 1 – cos²θ
= 1 – \(\left(-\frac{3}{5}\right)^{2}\)
= 1 – \(\frac{9}{25}=\frac{16}{25}\)
∴ sin θ = ± \(\frac{4}{5}\)
Since 180° < 0 < 270°,
θ lies in the 3rd quadrant.
∴ sin θ < 0

Since A lies in the 2nd quadrant,
tan A < 0

iii. Given, cot x = \(\frac{3}{4}\)
We know that,
cosec² x = 1 + cot² x
= 1 + \(\left(\frac{3}{4}\right)^{2}=1+\frac{9}{16}=\frac{25}{16}\)
∴ cosec x = ± \(\frac{5}{4}\)
Since x lies in the 3rd quadrant, cosec x < 0

iv. Given, tan x = \(-\frac{5}{12}\)
sec² x = 1 + tan²
= 1 + \(\left(-\frac{5}{12}\right)^{2}\)
= 1 + \(\frac{25}{144}=\frac{169}{144}\)
∴ sec x = ± \(\frac{13}{12}\)
Since x lies in the 4th quadrant,
sec x > 0