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Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Line and Plane Ex 6.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Ex 6.4

Question 1.
Find the angle between planes \(\bar{r} \cdot(\hat{i}+\hat{j}+2 \hat{k})\) = 13 and \(\bar{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 31 .
Solution:
The acute angle θ between the planes \(\bar{r} \cdot \bar{n}_{1}\) = d₁ and \(\bar{r} \cdot \bar{n}_{2}\) = d₂ is given by

Question 2.
Find the acute angle between the line \(\bar{r} \cdot(\hat{i}+2 \hat{j}+2 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}-6 \hat{k})\) and the plane \(\bar{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 0
Solution:
The acute angle θ between the line \(\bar{r}=\bar{a}+\lambda \bar{b}\) and the plane \(\bar{r} \cdot \bar{n}\) = d is given by

Question 3.
Show that lines \(\bar{r}=(2 \hat{j}-3 \hat{k})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k})\) and \(\bar{r}=(2 \hat{i}+6 \hat{j}+3 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+4 \hat{k})\) are coplanar. Find the equation of the plane determined by them.
Solution:

= 2(-1) + 6(2) + 3(-1)
= -2 + 12 – 3 = 7
∴ \(\bar{a}_{1} \cdot\left(\bar{b}_{1} \times \bar{b}_{2}\right)=\bar{a}_{2} \cdot\left(\bar{b}_{1} \times \bar{b}_{2}\right)\)
Hence, the given lines are coplanar.
The plane determined by these lines is given by
∴ \(\bar{r} \cdot\left(\overline{b_{1}} \times \overline{b_{2}}\right)=\overline{a_{1}} \cdot\left(\overline{b_{1}} \times \overline{b_{2}}\right)\)
i.e. \(\bar{r} \cdot(-\hat{i}+2 \hat{j}-\hat{k})\)
Hence, the given lines are coplanar and the equation of the plane determined by these lines is
\(\bar{r} \cdot(-\hat{i}+2 \hat{j}-\hat{k})\) = 7

Question 4.
Find the distance of the point \(4 \hat{i}-3 \hat{j}+\hat{k}\) from the plane \(\bar{r} \cdot(2 \hat{i}+3 \hat{j}-6 \hat{k})\) = 21 .
Solution:
The distance of the point A(\(\bar{a}\)) from the plane \(\bar{r} \cdot \bar{n}=p\) is given by d = \(\frac{|\bar{a} \cdot \bar{n}-p|}{|n|}\) …(1)
Here, \(\bar{a}=4 \hat{i}-3 \hat{j}+\hat{k}\), \(\bar{n}=2 \hat{i}+3 \hat{j}-6 \hat{k}\), p = 21
∴ \(\bar{a} \cdot \bar{n}\) = \((4 \hat{i}-3 \hat{j}+\hat{k}) \cdot(2 \hat{i}+3 \hat{j}-6 \hat{k})\)
= (4)(2) + (-3)(3) + (1)(-6)
= 8 – 9 – 6 = -7
Also, \(\sqrt{2^{2}+3^{2}+(-6)^{2}}=\sqrt{49}\) = 7
∴ from (1), the required distance
= \(\frac{|-7-21|}{7}\) = 4units

Question 5.
Find the distance of the point (1, 1, -1) from the plane 3x + 4y – 12z + 20 = 0.
Solution:
The distance of the point (x₁, y₁, z₁) from the plane ax + by + cz + d = 0 is \(\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|\)
∴ the distance of the point (1, 1, -1) from the plane 3x + 4y – 12z + 20 = 0 is \(\left|\frac{3(1)+4(1)-12(-1)+20}{\sqrt{3^{2}+4^{2}+(-12)^{2}}}\right|\)
= \(\left|\frac{3+4+12+20}{\sqrt{9+16+144}}\right|=\frac{39}{\sqrt{169}}\)
= \(\frac{39}{13}\) = 3units

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Ex 4.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.3

Question 1.
Find the joint equation of the pair of lines:
(i) Through the point (2, -1) and parallel to lines represented by 2x² + 3xy – 9y² = 0
Solution:
The combined equation of the given lines is
2x² + 3 xy – 9y² = 0
i.e. 2x² + 6xy – 3xy – 9y² = 0
i.e. 2x(x + 3y) – 3y(x + 3y) = 0
i.e. (x + 3y)(2x – 3y) = 0
∴ their separate equations are
x + 3y = 0 and 2x – 3y = 0
∴ their slopes are m₁ = \(\frac{-1}{3}\) and m₂ = \(\frac{-2}{-3}=\frac{2}{3}\).
The slopes of the lines parallel to these lines are m₁ and m₂, i.e. \(-\frac{1}{3}\) and \(\frac{2}{3}\).
∴ the equations of the lines with these slopes and through the point (2, -1) are
y + 1 = \(-\frac{1}{3}\) (x – 2) and y + 1 = \(\frac{2}{3}\)(x – 2)
i.e. 3y + 3= -x + 2 and 3y + 3 = 2x – 4
i.e. x + 3y + 1 = 0 and 2x – 3y – 7 = 0
∴ the joint equation of these lines is
(x + 3y + 1)(2x – 3y – 7) = 0
∴ 2x² – 3xy – 7x + 6xy – 9y² – 21y + 2x – 3y – 7 = 0
∴ 2x² + 3xy – 9y² – 5x – 24y – 7 = 0.

(ii) Through the point (2, -3) and parallel to lines represented by x² + xy – y² = 0
Solution:
Comparing the equation
x² + xy – y² = 0 … (1)
with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = 1, b = -1
Let m₁ and m₂ be the slopes of the lines represented by (1).

The slopes of the lines parallel to these lines are m₁ and m₂.
∴ the equations of the lines with these slopes and through the point (2, -3) are
y + 3 = m₁(x – 2) and y + 3 = m₂(x – 2)
i.e. m₁(x – 2) – (y + 3) = 0 and m₂(x – 2) – (y + 3) = 0
∴ the joint equation of these lines is
[m₁(x – 2) – (y + 3)][m₂(x – 2) – (y + 3)] = 0
∴ m₁m₂(x – 2)² – m₁(x – 2)(y + 3) – m₂(x – 2)(y + 3) + (y + 3)² = o
∴ m₁m₂(x – 2)² – (m₁ + m₂)(x – 2)(y + 3) + (y + 3)³ = 0
∴ -(x – 2)² – (x – 2)(y + 3) + (y + 3)² = 0 …… [By (2)]
∴ (x – 2)² + (x – 2)(y + 3) – (y + 3)² = 0
∴ (x² – 4x + 4) + (xy + 3x – 2y – 6) – (y² + 6y + 9) = 0
∴ x² – 4x + 4 + xy + 3x – 2y – 6 – y² – 6y – 9 = 0
∴ x² + xy – y² – x – 8y – 11 = 0.

Question 2.
Show that equation x² + 2xy+ 2y² + 2x + 2y + 1 = 0 does not represent a pair of lines.
Solution:
Comparing the equation
x² + 2xy + 2y² + 2x + 2y + 1 = 0 with
ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get,
a = 1, h = 1, b = 2, g = 1, f = 1, c = 1.
The given equation represents a pair of lines, if
D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|\) = 0 and h² – ab ≥ 0
Now, D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 1
\end{array}\right|\)
= 1 (2 – 1) – 1(1 – 1) + 1 (1 – 2)
= 1 – 0 – 1 = 0
and h² – ab = (1)² – 1(2) = -1 < 0
∴ given equation does not represent a pair of lines.

Question 3.
Show that equation 2x² – xy – 3y² – 6x + 19y – 20 = 0 represents a pair of lines.
Solution:
Comparing the equation
2x² – xy – 3y² – 6x + 19y – 20 = 0
with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get,
a = 2, h = \(-\frac{1}{2}\), b = -3, g = -3, f = \(\frac{19}{2}\), c = -20.
∴ D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{rrr}
2 & -\frac{1}{2} & -3 \\
-\frac{1}{2} & -3 & \frac{19}{2} \\
-3 & \frac{19}{2} & -20
\end{array}\right|\)
Taking \(\frac{1}{2}\) common from each row, we get,
D = \(\frac{1}{8}\left|\begin{array}{rrr}
4 & -1 & -6 \\
-1 & -6 & 19 \\
-6 & 19 & -40
\end{array}\right|\)
= \(\frac{1}{8}\)[4(240 – 361) + 1(40 + 114) – 6(-19 – 36)]
= \(\frac{1}{8}\)[4(-121) + 154 – 6(-55)]
= \(\frac{11}{8}\)[4(-11) + 14 – 6(-5)]
= \(\frac{1}{8}\)(-44 + 14 + 30) = 0
Also h² – ab = \(\left(-\frac{1}{2}\right)^{2}\) – 2(-3) = \(\frac{1}{4}\) + 6 = \(\frac{25}{4}\) > 0
∴ the given equation represents a pair of lines.

Question 4.
Show the equation 2x² + xy – y² + x + 4y – 3 = 0 represents a pair of lines. Also find the acute angle between them.
Solution:
Comparing the equation
2x² + xy — y² + x + 4y — 3 = 0 with
ax² + 2hxy + by² + 2gx + 2fy + c — 0, we get,
a = 2, h = \(\frac{1}{2}\), b = -1, g = \(\frac{1}{2}\), f = 2, c = – 3.
∴ D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{lrr}
2 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & -1 & 2 \\
\frac{1}{2} & 2 & -3
\end{array}\right|\)
Taking \(\frac{1}{2}\) common from each row, we get,
D = \(\frac{1}{8}\left|\begin{array}{rrr}
4 & 1 & 1 \\
1 & -2 & 4 \\
1 & 4 & -6
\end{array}\right|\)
= \(\frac{1}{8}\)[4(12 —16) — 1( —6 — 4) + 1(4 + 2)]
= \(\frac{1}{8}\)[4( – 4) – 1(-10) + 1(6)]
= \(\frac{1}{8}\)(—16 + 10 + 6) = 0
Also, h² – ab = \(\left(\frac{1}{2}\right)^{2}\) – 2(-1) = \(\frac{1}{4}\) + 2 = \(\frac{9}{4}\) > 0
∴ the given equation represents a pair of lines. Let θ be the acute angle between the lines
∴ tan θ = \(\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|\)

Question 5.
Find the separate equation of the lines represented by the following equations :
(i) (x – 2)² – 3(x – 2)(y + 1) + 2(y + 1)² = 0
Solution:
(x – 2)² – 3(x – 2)(y + 1) + 2(y + 1)2 = 0
∴ (x – 2)² – 2(x – 2)(y + 1) – (x – 2)(y + 1) + 2(y + 1)² = 0
∴ (x – 2) [(x – 2) – 2(y + 1)] – (y + 1)[(x – 2) – 2(y + 1)] = 0
∴ (x – 2)(x – 2 – 2y – 2) – (y + 1)(x – 2 – 2y – 2) = 0
∴ (x – 2)(x – 2y – 4) – (y + 1)(x – 2y – 4) = 0
∴ (x – 2y – 4)(x – 2 – y – 1) = 0
∴ (x – 2y – 4)(x – y – 3) = 0
∴ the separate equations of the lines are
x – 2y – 4 = 0 and x – y – 3 = 0.
Alternative Method :
(x – 2)² – 3(x – 2)(y + 1) + 2(y + 1)² = 0 … (1)
Put x – 2 = X and y + 1 = Y
∴ (1) becomes,
X² – 3XY + 2Y² = 0
∴ X² – 2XY – XY + 2Y² = 0
∴ X(X – 2Y) – Y(X – 2Y) = 0
∴ (X – 2Y)(X – Y) = 0
∴ the separate equations of the lines are
∴ X – 2Y = 0 and X – Y = 0
∴ (x – 2) – 2(y + 1) = 0 and (x – 2) – (y +1) = 0
∴ x – 2y – 4 = 0 and x – y – 3 = 0.

(ii) 10(x + 1)² + (x + 1)( y – 2) – 3(y – 2)² = 0
Solution:
10(x + 1)² + (x + 1)( y – 2) – 3(y – 2)² = 0 …(1)
Put x + 1 = X and y – 2 = Y
∴ (1) becomes
10x² + xy – 3y² = 0
10x² + 6xy – 5xy – 3y² = 0
2x(5x + 3y) – y(5x + 3y) = 0
(2x – y)(5x + 3y) = 0
5x + 3y = 0 and 2x – y = 0
5x + 3y = 0
5(x + 1) + 3(y – 2) = 0
5x + 5 + 3y – 6 = 0
∴ 5x + 3y – 1 = 0
2x – y = 0
2(x + 1) – (y – 2) = 0
2x + 2 – y + 2 = 0
∴ 2x – y + 4 = 0

Question 6.
Find the value of k if the following equations represent a pair of lines :
(i) 3x² + 10xy + 3y² + 16y + k = 0
Solution:
Comparing the given equation with
ax² + 2hxy + by² + 2gx + 2fy + c = 0,
we get, a = 3, h = 5, b = 3, g = 0, f= 8, c = k.
Now, given equation represents a pair of lines.
∴ abc + 2fgh – af² – bg² – ch² = 0
∴ (3)(3)(k) + 2(8)(0)(5) – 3(8)² – 3(0)² – k(5)² = 0
∴ 9k + 0 – 192 – 0 – 25k = 0
∴ -16k – 192 = 0
∴ – 16k = 192
∴ k= -12.

(ii) kxy + 10x + 6y + 4 = 0
Solution:
Comparing the given equation with
ax² + 2 hxy + by² + 2gx + 2fy + c = 0,
we get, a = 0, h = \(\frac{k}{2}\), b = 0, g = 5, f = 3, c = 4
Now, given equation represents a pair of lines.
∴ abc + 2fgh – af² – bg² – ch² = 0
∴ (0)(0)(4) + 2(3)(5)\(\left(\frac{k}{2}\right)\) – 0(3)² – 0(5)² – 4\(\left(\frac{k}{2}\right)^{2}\) = 0
∴ 0 + 15k – 0 – 0 – k² = 0
∴ 15k – k² = 0
∴ -k(k – 15) = 0
∴ k = 0 or k = 15.
If k = 0, then the given equation becomes
10x + 6y + 4 = 0 which does not represent a pair of lines.
∴ k ≠ o
Hence, k = 15.

(iii) x² + 3xy + 2y² + x – y + k = 0
Solution:
Comparing the given equation with
ax² + 2hxy + by² + 2gx + 2fy + c = 0,
we get, a = 1, h = \(\frac{3}{2}\), b = 2, g = \(\frac{1}{2}\), f= \(-\frac{1}{2}\), c = k.
Now, given equation represents a pair of lines.


∴ 2(8k – 1) – 3(6k + 1) + 1(-3 – 4) = 0
∴ 16k – 2 – 18k – 3 – 7 = 0
∴ -2k – 12 = 0
∴ -2k = 12 ∴ k = -6.

Question 7.
Find p and q if the equation px² – 8xy + 3y² + 14x + 2y + q = 0 represents a pair of perpendicular lines.
Solution:
The given equation represents a pair of lines perpendicular to each other
∴ (coefficient of x²) + (coefficient of y²) = 0
∴ p + 3 = 0 p = -3
With this value of p, the given equation is
– 3x² – 8xy + 3y² + 14x + 2y + q = 0.
Comparing this equation with
ax² + 2hxy + by² + 2gx + 2fy + c = 0, we have,
a = -3, h = -4, b = 3, g = 7, f = 1 and c = q.
D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{rrr}
-3 & -4 & 7 \\
-4 & 3 & 1 \\
7 & 1 & q
\end{array}\right|\)
= -3(3q – 1) + 4(-4q – 7) + 7(-4 – 21)
= -9q + 3 – 16q – 28 – 175
= -25q – 200= -25(q + 8)
Since the given equation represents a pair of lines, D = 0
∴ -25(q + 8) = 0 ∴ q= -8.
Hence, p = -3 and q = -8.

Question 8.
Find p and q if the equation 2x² + 8xy + py² + qx + 2y – 15 = 0 represents a pair of parallel lines.
Solution:
The given equation is
2x² + 8xy + py² + qx + 2y – 15 = 0
Comparing it with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get,
a = 2, h = 4, b = p, g = \(\frac{q}{2}\), f = 1, c = – 15
Since the lines are parallel, h² = ab
∴ (4)² = 2p ∴ P = 8
Since the given equation represents a pair of lines

i.e. – 242 + 240 + 2q + 2q – 2q² = 0
i.e. -2q² + 4q – 2 = 0
i.e. q² – 2q + 1 = 0
i.e. (q – 1)² = 0 ∴ q – 1 = 0 ∴ q = 1.
Hence, p = 8 and q = 1.

Question 9.
Equations of pairs of opposite sides of a parallelogram are x² – 7x+ 6 = 0 and y² – 14y + 40 = 0. Find the joint equation of its diagonals.
Solution:
Let ABCD be the parallelogram such that the combined equation of sides AB and CD is x² – 7x + 6 = 0 and the combined equation of sides BC and AD is y² – 14y + 40 = 0.
The separate equations of the lines represented by x² – 7x + 6 = 0, i.e. (x – 1)(x – 6) = 0 are x – 1 = 0 and x – 6 = 0.
Let equation of the side AB be x – 1 = 0 and equation of side CD be x – 6 = 0.
The separate equations of the lines represented by y² – 14y + 40 = 0, i.e. (y – 4)(y – 10) = 0 are y – 4 = 0 and y – 10 = 0.
Let equation of the side BC be y – 4 = 0 and equation of side AD be y – 10 = 0.

Coordinates of the vertices of the parallelogram are A(1, 10), B(1, 4), C(6, 4) and D(6, 10).
∴ equation of the diagonal AC is
\(\frac{y-10}{x-1}\) = \(\frac{10-4}{1-6}\) = \(\frac{6}{-5}\)
∴ -5y + 50 = 6x – 6
∴ 6x + 5y – 56 = 0
and equation of the diagonal BD is
\(\frac{y-4}{x-1}\) = \(\frac{4-10}{1-6}\) = \(\frac{-6}{-5}\) = \(\frac{6}{5}\)
∴ 5y – 20 = 6x – 6
∴ 6x – 5y + 14 = 0
Hence, the equations of the diagonals are 6x + 5y – 56 = 0 and 6x – 5y + 14 = 0.
∴ the joint equation of the diagonals is (6x + 5y – 56)(6x – 5y + 14) = 0
∴ 36x² – 30xy + 84x + 30xy – 25y² + 70y – 336x + 280y – 784 = 0
∴ 36x² – 25y² – 252x + 350y – 784 = 0.

Question 10.
∆OAB is formed by lines x² – 4xy + y² = 0 and the line 2x + 3y – 1 = 0. Find the equation of the median of the triangle drawn from O.
Solution:

Let D be the midpoint of seg AB where A is (x₁, y₁) and B is (x₂, y₂).
Then D has coordinates \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\).
The joint (combined) equation of the lines OA and OB is x² – 4xy + y² = 0 and the equation of the line AB is 2x + 3y – 1 = 0.
∴ points A and B satisfy the equations 2x + 3y – 1 = 0
and x² – 4xy + y² = 0 simultaneously.
We eliminate x from the above equations, i.e.,
put x = \(\frac{1-3 y}{2}\) in the equation x² – 4xy + y² = 0, we get,
∴ \(\left(\frac{1-3 y}{2}\right)^{2}\) – 4\(\left(\frac{1-3 y}{2}\right)\)y + y² = 0
∴ (1 – 3y)² – 8(1 – 3y)y + 4y² = 0
∴1 – 6y + 9y² – 8y + 24y² + 4y² = 0
∴ 37y² – 14y + 1 = 0
The roots y₁ and y₂ of the above quadratic equation are the y-coordinates of the points A and B.
∴ y₁ + y₂ = \(\frac{-b}{a}=\frac{14}{37}\)
∴ y-coordinate of D = \(\frac{y_{1}+y_{2}}{2}=\frac{7}{37}\).
Since D lies on the line AB, we can find the x-coordinate of D as
2x + 3\(\left(\frac{7}{37}\right)\) – 1 = 0
∴ 2x = 1 – \(\frac{21}{37}=\frac{16}{37}\)
∴ x = \(\frac{8}{37}\)
∴ D is (8/37, 7/37)
∴ equation of the median OD is \(\frac{x}{8 / 37}=\frac{y}{7 / 37}\),
i.e., 7x – 8y = 0.

Question 11.
Find the co-ordinates of the points of intersection of the lines represented by x² – y² – 2x + 1 = 0.
Solution:
Consider, x² – y² – 2x + 1 = 0
∴ (x² – 2x + 1) – y² = 0
∴ (x – 1)² – y² = 0
∴ (x – 1 + y)(x – 1 – y) = 0
∴ (x + y – 1)(x – y – 1) = 0
∴ separate equations of the lines are
x + y – 1 = 0 and x – y +1 = 0.
To find the point of intersection of the lines, we have to solve
x + y – 1 = 0 … (1)
and x – y + 1 = 0 … (2)
Adding (1) and (2), we get,
2x = 0 ∴ x = 0
Substituting x = 0 in (1), we get,
0 + y – 1 = 0 ∴ y = 1
∴ coordinates of the point of intersection of the lines are (0, 1).

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Ex 4.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.2

Question 1.
Show that lines represented by 3x² – 4xy – 3y² = 0 are perpendicular to each other.
Solution:
Comparing the equation 3x² – 4 xy – 3y² = 0 with ax² + 2hxy + by² = 0, we get, a = 3, 2h = -4, b = -3 Since a + b = 3 + (-3) = 0, the lines represented by 3x² – 4xy – 3y² = 0 are perpendicular to each other.

Question 2.
Show that lines represented by x² + 6xy + gy²= 0 are coincident.
The question is modified.
Show that lines represented by x² + 6xy + 9y²= 0 are coincident.
Solution:
Comparing the equation x² + 6xy + 9y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = 6, i.e. h = 3 and b = 9
Since h² – ab = (3)² – 1(9)
= 9 – 9 = 0, .
the lines represented by x² + 6xy + 9y² = 0 are coincident.

Question 3.
Find the value of k if lines represented by kx² + 4xy – 4y² = 0 are perpendicular to each other.
Solution:
Comparing the equation kx² + 4xy – 4y² = 0 with ax² + 2hxy + by² = 0, we get,
a = k, 2h = 4, b = -4
Since lines represented by kx² + 4xy – 4y² = 0 are perpendicular to each other,
a + b = 0
∴ k – 4 = 0 ∴ k = 4.

Question 4.
Find the measure of the acute angle between the lines represented by:
(i) 3x² – 4\(\sqrt {3}\)xy + 3y² = 0
Solution:
Comparing the equation 3x² – 4\(\sqrt {3}\)xy + 3y² = 0 with
ax² + 2hxy + by² = 0, we get,
a = 3, 2h = -4\(\sqrt {3}\), i.e. h = -24\(\sqrt {3}\) and b = 3
Let θ be the acute angle between the lines.

∴ θ = 30°.

(ii) 4x² + 5xy + y² = 0
Solution:
Comparing the equation 4x² + 5xy + y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 4, 2h = 5, i.e. h = \(\frac{5}{2}\) and b = 1.
Let θ be the acute angle between the lines.

(iii) 2x² + 7xy + 3y² = 0
Solution:
Comparing the equation
2x² + 7xy + 3y² = 0 with
ax² + 2hxy + by² = 0, we get,
a = 2, 2h = 7 i.e. h = \(\frac{7}{2}\) and b = 3
Let θ be the acute angle between the lines.


tanθ = 1
∴ θ = tan 1 = 45°
∴ θ = 45°

(iv) (a² – 3b²)x² + 8abxy + (b² – 3a²)y² = 0
Solution:
Comparing the equation
(a² – 3b²)x² + 8abxy + (b² – 3a²)y² = 0, with
Ax² + 2Hxy + By² = 0, we have,
A = a² – 3b², H = 4ab, B = b² – 3a².
∴ H² – AB = 16a²b² – (a² – 3b²)(b² – 3a²)
= 16a²b² + (a² – 3b²)(3a² – b²)
= 16a²b² + 3a⁴ – 10a²b² + 3b⁴
= 3a⁴ + 6a²b² + 3b⁴
= 3(a⁴ + 2a²b² + b⁴)
= 3 (a² + b²)²
∴ \(\sqrt{H^{2}-A B}\) = \(\sqrt {3}\) (a² + b²)
Also, A + B = (a² – 3b²) + (b² – 3a²)
= -2 (a² + b²)
If θ is the acute angle between the lines, then
tan θ = \(\left|\frac{2 \sqrt{H^{2}-A B}}{A+B}\right|=\left|\frac{2 \sqrt{3}\left(a^{2}+b^{2}\right)}{-2\left(a^{2}+b^{2}\right)}\right|\)
= \(\sqrt {3}\) = tan 60°
∴ θ = 60°

Question 5.
Find the combined equation of lines passing through the origin each of which making an angle of 30° with the line 3x + 2y – 11 = 0
Solution:
The slope of the line 3x + 2y – 11 = 0 is m₁ = \(-\frac{3}{2}\) .
Let m be the slope of one of the lines making an angle of 30° with the line 3x + 2y – 11 = 0.
The angle between the lines having slopes m and m1 is 30°.

On squaring both sides, we get,
\(\frac{1}{3}=\frac{(2 m+3)^{2}}{(2-3 m)^{2}}\)
∴ (2 – 3m)² = 3 (2m + 3)²
∴ 4 – 12m + 9m² = 3(4m² + 12m + 9)
∴ 4 – 12m + 9m² = 12m² + 36m + 27
3m² + 48m + 23 = 0
This is the auxiliary equation of the two lines and their joint equation is obtained by putting m = \(\frac{y}{x}\).
∴ the combined equation of the two lines is
3\(\left(\frac{y}{x}\right)^{2}\) + 48\(\left(\frac{y}{x}\right)\) + 23 = 0
∴ \(\frac{3 y^{2}}{x^{2}}+\frac{48 y}{x}\) + 23 = 0
∴ 3y² + 48xy + 23x² = 0
∴ 23x² + 48xy + 3y² = 0.

Question 6.
If the angle between lines represented by ax² + 2hxy + by² = 0 is equal to the angle between lines represented by 2x² – 5xy + 3y² = 0 then show that 100(h² – ab) = (a + b)².
Solution:
The acute angle θ between the lines ax² + 2hxy + by² = 0 is given by
tan θ = \(\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|\) ..(1)
Comparing the equation 2x² – 5xy + 3y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 2, 2h= -5, i.e. h = \(-\frac{5}{2}\) and b = 3
Let ∝ be the acute angle between the lines 2x² – 5xy + 3y² = 0.

This is the required condition.

Question 7.
Find the combined equation of lines passing through the origin and each of which making angle 60° with the Y- axis.
Solution:

Let OA and OB be the lines through the origin making an angle of 60° with the Y-axis.
Then OA and OB make an angle of 30° and 150° with the positive direction of X-axis.
∴ slope of OA = tan 30° = \(\frac{1}{\sqrt{3}}\)
∴ equation of the line OA is
y = \(\frac{1}{\sqrt{3}}\) = x, i.e. x – \(\sqrt {3}\)y = 0
Slope of OB = tan 150° = tan (180° – 30°)
= tan 30° = \(-\frac{1}{\sqrt{3}}\)
∴ equation of the line OB is
y = \(-\frac{1}{\sqrt{3}}\)x, i.e. x + \(\sqrt {3}\) y = 0
∴ required combined equation is
(x – \(\sqrt {3}\)y)(x + \(\sqrt {3}\)y) = 0
i.e. x² – 3y² = 0.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Ex 4.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.1

Question 1.
Find the combined equation of the following pairs of lines:
(i) 2x + y = 0 and 3x – y = 0
Solution:
The combined equation of the lines 2x + y = 0 and 3x – y = 0 is
(2x + y)( 3x – y) = 0
∴ 6x² – 2xy + 3xy – y² = 0
∴ 6x² – xy – y² = 0.

(ii) x + 2y – 1 = 0 and x – 3y + 2 = 0
Solution:
The combined equation of the lines x + 2y – 1 = 0 and x – 3y + 2 = 0 is
(x + 2y – 1)(x – 3y + 2) = 0
∴ x² – 3xy + 2x + 2xy – 6y² + 4y – x + 3y – 2 = 0
∴ x² – xy – 6y² + x + 7y – 2 = 0.

(iii) Passing through (2, 3) and parallel to the co-ordinate axes.
Solution:
Equations of the coordinate axes are x = 0 and y = 0.
∴ the equations of the lines passing through (2, 3) and parallel to the coordinate axes are x = 2 and
i.e. x – 2 = 0 and y – 3 = 0.
∴ their combined equation is
(x – 2)(y – 3) = 0.
∴ xy – 3x – 2y + 6 = 0.

(iv) Passing through (2, 3) and perpendicular to lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0
Solution:
Let L₁ and L₂ be the lines passing through the point (2, 3) and perpendicular to the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0 respectively.
Slopes of the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0 are \(\frac{-3}{2}\) and \(\frac{-1}{-3}=\frac{1}{3}\) respectively.
∴ slopes of the lines L₁ and L₂ are \(\frac{2}{3}\) and -3 respectively.
Since the lines L₁ and L₂ pass through the point (2, 3), their equations are
y – 3 = \(\frac{2}{3}\)(x – 2) and y – 3 = -3 (x – 2)
∴ 3y – 9 = 2x – 4 and y – 3= -3x + 6
∴ 2x – 3y + 5 = 0 and 3x – y – 9 = 0
∴ their combined equation is
(2x – 3y + 5)(3x + y – 9) = 0
∴ 6x² + 2xy – 18x – 9xy – 3y² + 27y + 15x + 5y – 45 = 0
∴ 6x² – 7xy – 3y² – 3x + 32y – 45 = 0.

(v) Passsing through (-1, 2),one is parallel to x + 3y – 1 = 0 and the other is perpendicular to 2x – 3y – 1 = 0.
Solution:
Let L₁ be the line passing through (-1, 2) and parallel to the line x + 3y – 1 = 0 whose slope is –\(\frac{1}{3}\).
∴ slope of the line L₁ is –\(\frac{1}{3}\)
∴ equation of the line L₁ is
y – 2 = –\(\frac{1}{3}\)(x + 1)
∴ 3y – 6 = -x – 1
∴ x + 3y – 5 = 0
Let L₂ be the line passing through (-1, 2) and perpendicular to the line 2x – 3y – 1 = 0
whose slope is \(\frac{-2}{-3}=\frac{2}{3}\).
∴ slope of the line L₂ is –\(\frac{3}{2}\)
∴ equation of the line L₂ is
y – 2= –\(\frac{3}{2}\)(x + 1)
∴ 2y – 4 = -3x – 3
∴ 3x + 2y – 1 = 0
Hence, the equations of the required lines are
x + 3y – 5 = 0 and 3x + 2y – 1 = 0
∴ their combined equation is
(x + 3y – 5)(3x + 2y – 1) = 0
∴ 3x² + 2xy – x + 9xy + 6y² – 3y – 15x – 10y + 5 = 0
∴ 3x² + 11xy + 6y² – 16x – 13y + 5 = 0

Question 2.
Find the separate equations of the lines represented by following equations:
(i) 3y² + 7xy = 0
Solution:
3y² + 7xy = 0
∴ y(3y + 7x) = 0
∴ the separate equations of the lines are y = 0 and 7x + 3y = 0.

(ii) 5x² – 9y² = 0
Solution:
5x² – 9y² = 0
∴ (\(\sqrt {5}\) x)² – (3y)² = 0
∴ (\(\sqrt {5}\)x + 3y)(\(\sqrt {5}\)x – 3y) = 0
∴ the separate equations of the lines are
\(\sqrt {5}\)x + 3y = 0 and \(\sqrt {5}\)x – 3y = 0.

(iii) x² – 4xy = 0
Solution:
x² – 4xy = 0
∴ x(x – 4y) = 0
∴ the separate equations of the lines are x = 0 and x – 4y = 0

(iv) 3x² – 10xy – 8y² = 0
Solution:
3x² – 10xy – 8y² = 0
∴ 3x² – 12xy + 2xy – 8y² = 0
∴ 3x(x – 4y) + 2y(x – 4y) = 0
∴ (x – 4y)(3x +2y) = 0
∴ the separate equations of the lines are x – 4y = 0 and 3x + 2y = 0.

(v) 3x² – \(2 \sqrt{3}\) xy – 3y² = 0
Solution:
3x² – 2\(\sqrt {3}\)xy – 3y2 = 0
∴ 3x² – 3\(\sqrt {3}\)xy + \(\sqrt {3}\)xy – 3y² = 0
∴ 3x(x – \(\sqrt {3}\)y) + \(\sqrt {3}\)y(x – \(\sqrt {3}\)y) = 0
∴ (x – \(\sqrt {3}\)y)(3x + \(\sqrt {3}\)y) = 0
∴ the separate equations of the lines are
∴ x – \(\sqrt {3}\)y = 0 and 3x + \(\sqrt {3}\)y = 0.

(vi) x² + 2(cosec ∝)xy + y² = 0
Solution:
x² + 2 (cosec ∝)xy – y² = 0
i.e. y² + 2(cosec∝)xy + x² = 0
Dividing by x², we get,

∴ the separate equations of the lines are
(cosec ∝ – cot ∝)x + y = 0 and (cosec ∝ + cot ∝)x + y = 0.

(vii) x² + 2xy tan ∝ – y² = 0
Solution:
x² + 2xy tan ∝ – y² = 0
Dividind by y²

The separate equations of the lines are
(sec∝ – tan ∝)x + y = 0 and (sec ∝ + tan ∝)x – y = 0

Question 3.
Find the combined equation of a pair of lines passing through the origin and perpendicular
to the lines represented by following equations :
(i) 5x² – 8xy + 3y² = 0
Solution:
Comparing the equation 5x² – 8xy + 3y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 5, 2h = -8, b = 3
Let m₁ and m₂ be the slopes of the lines represented by 5x² – 8xy + 3y² = 0.
∴ m₁ + m₂ = \(\frac{-2 h}{b}=\frac{8}{3}\)
amd m₁m₂ = \(\frac{a}{b}=\frac{5}{3}\) …(1)
Now required lines are perpendicular to these lines
∴ their slopes are -1 /m₁ and -1/m₂ Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m₁y = -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁y) (x + m₂y) = 0
∴ x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x² + \(\frac{8}{3}\)xy + \(\frac{5}{3}\)y² = 0 … [By (1)]
∴ x² + 8xy + 5y\(\frac{8}{3}\) = 0

(ii) 5x² + 2xy – 3y² = 0
Solution:
Comparing the equation 5x² + 2xy – 3y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 5, 2h = 2, b = -3
Let m₁ and m₂ be the slopes of the lines represented by 5x² + 2xy – 3y² = 0
∴ m₁ + m₂ = \(\frac{-2 h}{b}=\frac{-2}{-3}=\frac{2}{3}\) and m₁m₂ = \(\frac{a}{b}=\frac{5}{-3}\) ..(1)
Now required lines are perpendicular to these lines
∴ their slopes are \(\frac{-1}{m_{1}}\) and \(\frac{-1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{\mathrm{~m}_{1}}\)x and y = \(\frac{-1}{\mathrm{~m}_{2}}\)x
i.e. m₁y = -x amd m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
∴ (x + m₁y)(x + m₂y) = 0
x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x² + \(\frac{2}{3}\)xy – \(\frac{5}{3}\)y = 0 …[By (1)]
∴ 3x² + 2xy – 5y² = 0

(iii) xy + y² = 0
Solution:
Comparing the equation xy + y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 0, 2h = 1, b = 1
Let m₁ and m₂ be the slopes of the lines represented by xy + y² = 0

Now required lines are perpendicular to these lines
∴ their slopes are \(\frac{-1}{m_{1}}\) and \(\frac{-1}{m_{2}}\).
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m₁y = -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁y) (x + m₂y) = 0
∴ x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x² – xy = 0.y² = 0 … [By (1)]
∴ x² – xy = 0.
Alternative Method :
Consider xy + y² = 0
∴ y(x + y) = 0
∴ separate equations of the lines are y = 0 and
3x² + 8xy + 5y² = 0.
x + y = 0.
Let m₁ and m₂ be the slopes of these lines.
Then m₁ = 0 and m₂ = -1
Now, required lines are perpendicular to these lines.
∴ their slopes are \(-\frac{1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since, m₁ = 0, \(-\frac{1}{m_{1}}\) does not exist.
Also, m₂ = -1, \(-\frac{1}{m_{2}}\) = 1
Since these lines are passing through the origin, their separate equations are x = 0 and y = x,
i.e. x – y = 0
∴ their combined equation is
x(x – y) = 0
x² – xy = 0.

(iv) 3x² – 4xy = 0
Solution:
Consider 3x² – 4xy = 0
∴ x(3x – 4y) = 0
∴ separate equations of the lines are x = 0 and 3x – 4y = 0.
Let m₁ and m₂ be the slopes of these lines.
Then m₁ does not exist and and m₁ = \(\frac{3}{4}\).
Now, required lines are perpendicular to these lines.
∴ their slopes are \(-\frac{1}{m_{1}}\) and \(-\frac{1}{m_{2}}\).
Since m₁ does not exist, \(-\frac{1}{m_{1}}\) = 0
Also m₂ = \(\frac{3}{4^{\prime}}-\frac{1}{m_{2}}=-\frac{4}{3}\)
Since these lines are passing through the origin, their separate equations are y = 0 and y = \(-\frac{4}{3}\)x,
i.e.   4x + 3y = 0
∴ their combined equation is
y(4x + 3y) = 0
∴ 4xy + 3y² = 0.

Question 4.
Find k if,
(i) the sum of the slopes of the lines represented by x² + kxy – 3y² = 0 is twice their product.
Solution:
Comparing the equation x² + kxy – 3y² = 0 with ax² + 2hxy + by² = 0, we get, a = 1, 2h = k, b = -3.
Let m₁ and m₂ be the slopes of the lines represented by x² + kxy – 3y² = 0.
∴ m₁ + m₂ = \(\frac{-2 h}{b}=-\frac{k}{(-3)}=\frac{k}{3}\)
and m₁m₂ = \(\frac{a}{b}=\frac{1}{(-3)}=-\frac{1}{3}\)
Now, m₁ + m₂ = 2(m₁m₂) ..(Given)
∴ \(\frac{k}{3}=2\left(-\frac{1}{3}\right)\) ∴ k = -2

(ii) slopes of lines represent by 3x² + kxy – y² = 0 differ by 4.
Solution:
(ii) Comparing the equation 3x² + kxy – y² = 0 with ax² + 2hxy + by² = 0, we get, a = 3, 2h = k, b = -1.
Let m₁ and m₂ be the slopes of the lines represented by 3x² + kxy – y² = 0.
∴ m₁ + m₂ = \(\frac{-2 h}{b}=-\frac{k}{-1}\) = k
and m₁₂ = \(\frac{a}{b}=\frac{3}{-1}\) = -3
∴ (m₁ – m₂)² = (m₁ + m₂)² – 4m₁m₂
= k² – 4 (-3)
= k² + 12 … (1)
But |m₁ – m₂| =4
∴ (m₁ – m₂)² = 16 … (2)
∴ from (1) and (2), k² + 12 = 16
∴ k² = 4 ∴ k= ±2.

(iii) slope of one of the lines given by kx² + 4xy – y² = 0 exceeds the slope of the other by 8.
Solution:
Comparing the equation kx² + 4xy – y² = 0 with ² + 2hxy + by² = 0, we get, a = k, 2h = 4, b = -1. Let m₁ and m₂ be the slopes of the lines represented by kx² + 4xy – y² = 0.
∴ m₁ + m₂ = \(\frac{-2 h}{b}=\frac{-4}{-1}\) = 4
and m₁m₂ = \(\frac{a}{b}=\frac{k}{-1}\) = -k
We are given that m₂ = m₁ + 8
m₁ + m₁ + 8 = 4
∴ 2m₁ = -4 ∴ m₁ = -2 … (1)
Also, m₁(m₁ + 8) = -k
(-2)(-2 + 8) = -k … [By(1)]
∴ (-2)(6) = -k
∴ -12= -k ∴ k = 12.

Question 5.
Find the condition that :
(i) the line 4x + 5y = 0 coincides with one of the lines given by ax² + 2hxy + by² = 0.
Solution:
The auxiliary equation of the lines represented by ax² + 2hxy + by² = 0 is bm² + 2hm + a = 0.
Given that 4x + 5y = 0 is one of the lines represented by ax² + 2hxy + by² = 0.
The slope of the line 4x + 5y = 0 is \(-\frac{4}{5}\).
∴ m = \(-\frac{4}{5}\) is a root of the auxiliary equation bm² + 2hm + a = 0.
∴ b\(\left(-\frac{4}{5}\right)^{2}\) + 2h\(\left(-\frac{4}{5}\right)\) + a = 0
∴ \(\frac{16 b}{25}-\frac{8 h}{5}\) + a = 0
∴ 16b – 40h + 25a = 0
∴ 25a + 16b = 40k.
This is the required condition.

(ii) the line 3x + y = 0 may be perpendicular to one of the lines given by ax² + 2hxy + by² = 0.
Solution:
The auxiliary equation of the lines represented by ax² + 2hxy + by² = 0 is bm² + 2hm + a = 0.
Since one line is perpendicular to the line 3x + y = 0
whose slope is \(-\frac{3}{1}\) = -3
∴ slope of that line = m = \(\frac{1}{3}\)
∴ m = \(\frac{1}{3}\)is the root of the auxiliary equation bm² + 2hm + a = 0.
∴ b\(\left(\frac{1}{3}\right)^{2}\) + 2h\(\left(\frac{1}{3}\right)\) + a = 0
∴ \(\frac{b}{9}+\frac{2 h}{3}\) + a = 0
∴ b + 6h + 9a = 0
∴ 9a + b + 6h = 0
This is the required condition.

Question 6.
If one of the lines given by ax² + 2hxy + by² = 0 is perpendicular to px + qy = 0 then show that ap² + 2hpq + bq² = 0.
Solution:
To prove ap² + 2hpq + bq² = 0.
Let the slope of the pair of straight lines ax² + 2hxy + by² = 0 be m₁ and m₂
Then, m₁ + m₂ = \(\frac{-2 h}{b}\) and m₁m₂ = \(\frac{a}{b}\)
Slope of the line px + qy = 0 is \(\frac{-p}{q}\)
But one of the lines of ax² + 2hxy + by² = 0 is perpendicular to px + qy = 0

⇒ bq² + ap² = -2hpq
⇒ ap² + 2hpq + bq² = 0

Question 7.
Find the combined equation of the pair of lines passing through the origin and making an equilateral triangle with the line y = 3.
Solution:
Let OA and OB be the lines through the origin making.an angle of 60° with the line y = 3.
∴ OA and OB make an angle of 60° and 120° with the positive direction of X-axis.
∴ slope of OA = tan60° = \(\sqrt {3}\)
∴ equation of the line OA is
y = \(\sqrt {3}\) x, i.e. \(\sqrt {3}\) x – y = 0

Slope of OB = tan 120° = tan (180° – 60°)
= -tan 60°= –\(\sqrt {3}\)
∴ equation of the line OB is
y = –\(\sqrt {3}\) x, i.e. \(\sqrt {3}\) x + y = 0
∴ required joint equation of the lines is
(\(\sqrt {3}\) x – y)(\(\sqrt {3}\) x + y) = 0
i.e. 3x² – y² = 0.

Question 8.
If slope of one of the lines given by ax² + 2hxy + by² = 0 is four times the other then show that 16h² = 25ab.
Solution:
Let m₁ and m₂ be the slopes of the lines given by ax² + 2hxy + by² = 0.
∴ m₁ + m₂ = \(-\frac{2 h}{b}\)
and m₁m₂ = \(\frac{a}{b}\)
We are given that m₂ = 4m₁

∴ 16h² = 25ab
This is the required condition.

Question 9.
If one of the lines given by ax² + 2hxy + by² = 0 bisects an angle between co-ordinate axes then show that (a + b) ² = 4h².
Solution:
The auxiliary equation of the lines given by ax² + 2hxy + by² = 0 is bm² + 2hm + a = 0.
Since one of the line bisects an angle between the coordinate axes, that line makes an angle of 45° or 135° with the positive direction of X-axis.
∴ slope of that line = tan45° or tan 135°
∴ m = tan45° = 1
or m = tan 135° = tan (180° – 45°)
= -tan 45°= -1
∴ m = ±1 are the roots of the auxiliary equation bm² + 2hm + a = 0.
∴ b(±1)² + 2h(±1) + a = 0
∴ b ± 2h + a = 0
∴ a + b = ±2h
∴ (a + b)² = 4h²
This is the required condition.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Miscellaneous Exercise 3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Trigonometric Functions Miscellaneous Exercise 3

I) Select the correct option from the given alternatives.
Question 1.
The principal of solutions equation sinθ = \(\frac{-1}{2}\) are ________.

Solution:
(b) \(\frac{7 \pi}{6}, \frac{11 \pi}{6}\)

Question 2.
The principal solution of equation cot θ = \(\sqrt {3}\) ___________.

Solution:
(a) \(\frac{\pi}{6}, \frac{7 \pi}{6}\)

Question 3.
The general solution of sec x = \(\sqrt {2}\) is __________.
(a) 2nπ ± \(\frac{\pi}{4}\), n ∈ Z
(b) 2nπ ± \(\frac{\pi}{2}\), n ∈ Z
(c) nπ ± \(\frac{\pi}{2}\), n ∈ Z
(d) 2nπ ± \(\frac{\pi}{3}\), n ∈ Z
Solution:
(a) 2nπ ± \(\frac{\pi}{4}\), n ∈ Z

Question 4.
If cos pθ = cosqθ, p ≠ q rhen ________.
(a) θ = \(\frac{2 n \pi}{p \pm q}\)
(b) θ = 2nπ
(c) θ = 2nπ ± p
(d) nπ ± q
Solution:
(a) θ = \(\frac{2 n \pi}{p \pm q}\)

Question 5.
If polar co-ordinates of a point are \(\left(2, \frac{\pi}{4}\right)\) then its cartesian co-ordinates are ______.
(a) (2, \(\sqrt {2}\) )
(b) (\(\sqrt {2}\), 2)
(c) (2, 2)
(d) (\(\sqrt {2}\) , \(\sqrt {2}\))
Solution:
(d) (\(\sqrt {2}\) , \(\sqrt {2}\))

Question 6.
If \(\sqrt {3}\) cosx – sin x = 1, then general value of x is _________.
(a) 2nπ ± \(\frac{\pi}{3}\)
(b) 2nπ ± \(\frac{\pi}{6}\)
(c) 2nπ ± \(\frac{\pi}{3}-\frac{\pi}{6}\)
(d) nπ + (-1)ⁿ\(\frac{\pi}{3}\)
Solution:
(c) 2nπ ± \(\frac{\pi}{3}-\frac{\pi}{6}\)

Question 7.
In ∆ABC if ∠A = 45°, ∠B = 60° then the ratio of its sides are _________.
(a) 2 : \(\frac{\pi}{2}\) : \(\frac{\pi}{3}\) + 1
(b) \(\frac{\pi}{2}\) : 2 : \(\frac{\pi}{3}\) + 1
(c) 2 \(\frac{\pi}{2}\) : \(\frac{\pi}{2}\) : \(\frac{\pi}{3}\)
(d) 2 : 2 \(\frac{\pi}{2}\) : \(\frac{\pi}{3}\) + 1
Solution:
(a) 2 : \(\frac{\pi}{2}\) : \(\frac{\pi}{3}\) + 1

Question 8.
In ∆ABC, if c² + a² – b² = ac, then ∠B = __________.
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{2}\)
(d) \(\frac{\pi}{6}\)
Solution:
(b) \(\frac{\pi}{3}\)

Question 9.
In ABC, ac cos B – bc cos A = ____________.
(a) a² – b²
(b) b² – c²
(c) c² – a²
(d) a² – b² – c²
Solution:
(a) a² – b²

Question 10.
If in a triangle, the are in A.P. and b : c = \(\sqrt {3}\) : \(\sqrt {2}\) then A is equal to __________.
(a) 30°
(b) 60°
(c) 75°
(d) 45°
Solution:
(c) 75°

Question 11.
cos^-1\(\left(\cos \frac{7 \pi}{6}\right)\) = ________.

Question 12.
The value of cot (tan^-1 2x + cot^-1 2x) is __________.
(a) 0
(b) 2x
(c) π + 2x
(d) π – 2x
Solution:
(a) 0

Question 13.
The principal value of sin^-1\(\left(-\frac{\sqrt{3}}{2}\right)\) is ____________.

Solution:
(d) \(-\frac{\pi}{3}\)

Question 14.
If sin^-1\(\frac{4}{5}\) + cos^-1\(\frac{,12}{13}\) = sin^-1 ∝, then ∝ = _____________.
(a) \(\frac{63}{65}\)
(b) \(\frac{62}{65}\)
(c) \(\frac{61}{65}\)
(d) \(\frac{60}{65}\)
Solution:
(a) \(\frac{63}{65}\)

Question 15.
If tan^-1(2x) + tan^-1(3x) = \(\frac{\pi}{4}\), then x = ________.
(a) -1
(b) \(\frac{1}{6}\)
(c) \(\frac{2}{6}\)
(d) \(\frac{3}{2}\)
Solution:
(b) \(\frac{1}{6}\)

Question 16.
2 tan^-1\(\frac{1}{3}\) + tan^-1\(\frac{1}{7}\) = ______.
(a) tan^-1\(\frac{4}{5}\)
(b) \(\frac{\pi}{2}\)
(c) 1
(d) \(\frac{\pi}{4}\)
Solution:
(d) \(\frac{\pi}{4}\)

Question 17.
tan (2 tan^-1\(\left(\frac{1}{5}\right)-\frac{\pi}{4}\)) = ______.
(a) \(\frac{17}{7}\)
(b) \(-\frac{17}{7}\)
(c) \(\frac{7}{17}\)
(d) \(-\frac{7}{17}\)
Solution:
(d) \(-\frac{7}{17}\)

Question 18.
The principal value branch of sec^-1 x is __________.

Solution:
(b) [0, π] – {\(\frac{\pi}{2}\)}

Question 19.
cos[tan^-1\(\frac{1}{3}\) + tan^-1\(\frac{1}{2}\)] = ________.
(a) \(\frac{1}{\sqrt{2}}\)
(b) \(\frac{\sqrt{3}}{2}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{\pi}{4}\)
Solution:
(a) \(\frac{1}{\sqrt{2}}\)

Question 20.
If tan θ + tan 2θ + tan 3θ = tan θ∙tan 2θ∙tan 3θ, then the general value of the θ is _______.
(a) nπ
(b) \(\frac{n \pi}{6}\)
(c) nπ ± \(\frac{n \pi}{4}\)
(d) \(\frac{n \pi}{2}\)
Solution:
(b) \(\frac{n \pi}{6}\)
[Hint: tan(A + B + C) = \(\frac{\tan A+\tan B+\tan C-\tan A \cdot \tan B \cdot \tan C}{1-\tan A \cdot \tan B-\tan B \cdot \tan C-\tan C \cdot \tan A}\)
Since , tan θ + tan 2θ + tan 3θ = tan θ ∙ tan 2θ ∙ tan 3θ,
we get, tan (θ + 2θ + 3θ) = θ
∴ tan6θ = 0
∴ 6θ = nπ, θ = \(\frac{n \pi}{6}\).]

Question 21.
If any ∆ABC, if a cos B = b cos A, then the triangle is ________.
(a) Equilateral triangle
(b) Isosceles triangle
(c) Scalene
(d) Right angled
Solution:
(b) Isosceles triangle

II: Solve the following
Question 1.
Find the principal solutions of the following equations :
(i) sin2θ = \(-\frac{1}{2}\)
Solution:
sin2θ = \(-\frac{1}{2}\)
Since, θ ∈ (0, 2π), 2∈ ∈ (0, 4π)

(ii) tan3θ = -1
Solution:
Since, θ ∈ (0, 2π), 3∈ ∈ (0, 6π)

… [∵ tan(π – θ) = tan(2π – θ) = tan(3π – θ)
= tan (4π – θ) = tan (5π – θ) = tan (6π – θ) = -tan θ]
∴ tan3θ = tan\(\frac{3 \pi}{4}\) = tan\(\frac{7 \pi}{4}\) = tan\(\frac{11 \pi}{4}\) = tan\(\frac{15 \pi}{4}\)

(iii) cotθ = 0
Solution:
cotθ = 0
Since θ ∈ (0, 2π),
cotθ = 0 = cot \(\frac{\pi}{2}\) = cot (π + \(\frac{\pi}{2}\) …[∵ cos(π + θ) = cotθ]
∴ cotθ = cot\(\frac{\pi}{2}\) = cot\(\frac{3 \pi}{2}\)
∴ θ = \(\frac{\pi}{2}\) or θ = \(\frac{3 \pi}{2}\)
Hence, the required principal solutions are \(\left\{\frac{\pi}{2}, \frac{3 \pi}{2}\right\}\)

Question 2.
Find the principal solutions of the following equations :
(i) sin2θ = \(-\frac{1}{\sqrt{2}}\)
Solution:

(ii) tan5θ = -1
Solution:

(iii) cot2θ = 0
Solution:

Question 3.
Which of the following equations have no solutions ?
(i) cos 2θ = \(\frac{1}{3}\)
Solution:
cos 2θ = \(\frac{1}{3}\)
Since \(\frac{1}{3}\) ≤ cosθ ≤ 1 for any θ
cos2θ = \(\frac{1}{3}\) has solution

(ii) cos² θ = -1
Solution:
cos2θ = -1
This is not possible because cos2θ ≥ 0 for any θ.
∴ cos2θ = -1 does not have any solution.

(iii) 2 sinθ = 3
Solution:
2 sin θ = 3 ∴ sin θ = \(\frac{3}{2}\)
This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 2 sin θ = 3 does not have any solution.

(iv) 3 sin θ = 5
Solution:
3 sin θ = 5
∴ sin θ = \(\frac{5}{3}\)
This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 3 sin θ = 5 does not have any solution.

Question 4.
Find the general solutions of the following equations :
(i) tanθ = \(-\sqrt {x}\)
Solution:
The general solution of tan θ = tan ∝ is
θ = nπ + ∝, n ∈ Z.
Now, tanθ = \(-\sqrt {x}\)
∴ tanθ = tan\(\frac{\pi}{3}\) …[∵ tan\(\frac{\pi}{3}\) = \(\sqrt {3}\)]
∴ tanθ = tan\(\left(\pi-\frac{\pi}{3}\right)\) …[∵ tan(π – θ) = -tanθ]
∴ tanθ = tan\(\frac{2 \pi}{3}\)
∴ the required general solution is
θ = nπ + \(\frac{2 \pi}{3}\), n ∈ Z.

(ii) tan²θ = 3
Solution:
The general solution of tan²θ = tan²∝ is
θ = nπ ± ∝, n ∈ Z.
Now, tan²θ = 3 = (\(\sqrt {x}\))²
∴ tan²θ = (tan\(\frac{\pi}{3}\))² …[∵ tan\(\frac{\pi}{3}\) = \(\sqrt {3}\)]
∴ tan²θ = tan²\(\frac{\pi}{3}\)
∴ the required general solution is
θ = nπ ± \(\frac{\pi}{3}\), n ∈ Z.

(iii) sin θ – cosθ = 1
Solution:
∴ cosθ – sin θ = -1

(iv) sin²θ – cos²θ = 1
Solution:
sin²θ – cos²θ = 1
∴ cos²θ – sin²θ = -1
∴ cos2θ = cosπ …(1)
The general solution of cos θ = cos ∝ is
θ = 2nπ ± ∝, n ∈ Z
∴ the general solution of (1) is given by
2θ = 2nπ ± π, n ∈ Z
∴ θ = nπ ± \(\frac{\pi}{2}\), n ∈ Z

Question 5.
In ∆ABC prove that cos \(\left(\frac{A-B}{2}\right)=\left(\frac{a+b}{c}\right)\) sin \(\frac{C}{2}\)
Solution:
By the sine rule,

Question 6.
With usual notations prove that \(\frac{\sin (A-B)}{\sin (A+B)}=\frac{a^{2}-b^{2}}{c^{2}}\).
Solution:
By the sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\) = k
∴ a = ksinA, b = ksinB, c = ksinC

Question 7.
In ∆ABC prove that (a – b)² 2cos²\(\frac{\mathrm{C}}{2}\) + (a + b)² sin²\(\frac{\mathrm{C}}{2}\) = c².
Solution:
LHS (a – b)² 2cos²\(\frac{\mathrm{C}}{2}\) + (a + b)² sin²\(\frac{\mathrm{C}}{2}\)
= (a² + b² – 2ab) cos²\(\frac{\mathrm{C}}{2}\) + (a² + b² + 2ab) sin\(\frac{\mathrm{C}}{2}\)²
= (a² + b²) cos²\(\frac{\mathrm{C}}{2}\) – 2ab cos²\(\frac{\mathrm{C}}{2}\) + (a² + b²) sin²\(\frac{\mathrm{C}}{2}\) + 2ab sin²\(\frac{\mathrm{C}}{2}\)
= (a² + b²) (cos²\(\frac{\mathrm{C}}{2}\) + sin²\(\frac{\mathrm{C}}{2}\)) – 2ab(cos²\(\frac{\mathrm{C}}{2}\) – sin²\(\frac{\mathrm{C}}{2}\))
= a² + b² – 2ab cos C
= c² = RHS.

Question 8.
In ∆ABC if cosA = sin B – cos C then show that it is a right angled triangle.
Solution:
cos A= sin B – cos C
∴ cos A + cos C = sin B

∴ A – C = B
∴ A = B + C
∴ A + B + C = 180° gives
A + A = 180°
∴ 2A = 180 ∴ A = 90°
∴ ∆ ABC is a rightangled triangle.

Question 9.
If \(\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}\) then show that a², b², c², are in A.P.
Solution:
By sine rule,
\(\frac{\sin \mathrm{A}}{a}\) = \(\frac{\sin \mathrm{B}}{b}\) = \(\frac{\sin \mathrm{C}}{c}\) = k
∴ sin A = ka, sin B = kb,sin C = kc
Now, \(\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}\)
∴ sinA∙sin(B – C) = sinC∙sin(A -B)
∴ sin [π – (B + C)] ∙ sin (B – C)
= sin [π – (A + B)]∙sin (A – B) … [∵ A + B + C = π]
∴ sin(B + C) ∙ sin(B – C) = sin (A + B) ∙ sin (A – B)
∴ sin²B – sin²C = sin²A – sin²B
∴ 2 sin²B = sin²A + sin²C
∴ 2k²b² = k²a² + k²c²
∴ 2b² = a² + c²
Hence, a², b², c² are in A.P.

Question 10.
Solve the triangle in which a = (\(\sqrt {3}\) + 1), b = (\(\sqrt {3}\) – 1) and ∠C = 60°.
Solution:
Given : a = \(\sqrt {3}\) + 1, b = \(\sqrt {3}\) – 1 and ∠C = 60°.
By cosine rule,
c² = a² + b² – 2ab cos C
= (\(\sqrt {3}\) + 1)² + (\(\sqrt {3}\) – 1)² – 2(\(\sqrt {3}\) + 1)(\(\sqrt {3}\) – 1)cos60°
= 3 + 1 + 2\(\sqrt {3}\) + 3+ 1 – 2\(\sqrt {3}\) – 2(3 – 1)\(\left(\frac{1}{2}\right)\)
= 8 – 2 = 6
∴ c = \(\sqrt {6}\) …[∵ c > 0)
By sine rule,

∴ sin A = sin 60° cos 45° + cos 60° sin 45°
and sin B = sin 60° cos 45° – cos 60° sin 45°
∴ sin A = sin (60° + 45°) – sin 105°
and sin B = sin (60° – 45°) = sin 15°
∴ A = 105° and B = 15°
Hence, A = 105°, B 15° and C = \(\sqrt {6}\) units.

Question 11.
In ∆ABC prove the following :
(i) a sin A – b sin B = c sin (A – B)
Solution:
By sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\) = k
∴ a = ksinA, b = ksinB, c = ksinC,
LHS = a sin A – b sinB
= ksinA∙sinA – ksinB∙sinB
= k (sin²A – sin²B)
= k (sin A + sin B)(sin A – sin B)

= k × sin (A + B) × sin (A – B)
= ksin(π – C)∙sin(A – B) … [∵ A + B + C = π]
= k sinC∙sin (A – B)
= c sin (A – B) = RHS.

(ii) \(\frac{c-b \cos A}{b-c \cos A}=\frac{\cos B}{\cos C}\).
Solution:

(iii) a² sin (B – C) = (b² – c²) sinA
Solution:
By sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\) = k
∴ a = ksinA, b = ksinB, c = ksinC
RHS = (b² – c²) sin A
= (k²sin²B – k²sin²C)sin A
= k²(sin²B – sin²C) sin A
= k²(sin B + sin C)(sin B – sin C) sin A

= k² × sin (B + C) × sin (B – C) × sin A
= k²∙sin(π – A)∙sin(B – C)∙sinA … [∵ A + B + C = π]
= k²sin A∙sin (B – C)∙sin A
= (k sin A)²∙sin (B – C)
= a²sin (B – C) = LHS.

(iv) ac cos B – bc cos A = (a² – b²).
Solution:
LHS = ac cos B – bc cos A
= ac\(\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)\) – bc\(\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)\)
=\(\frac{1}{2}\)(c² + a² – b²) – \(\frac{1}{2}\)(b² + c² – a²)
= \(\frac{1}{2}\)(c² + a² – b² – b² – c² + a²)
= \(\frac{1}{2}\)(2a² – 2b²) = a² – b2 = RHS.

(v) \(\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^{2}+b^{2}+c^{2}}{2 a b c}\) .
Solution:

(vi) \(\frac{\cos 2 \mathrm{~A}}{a^{2}}-\frac{\cos 2 \mathrm{~B}}{b^{2}}=\frac{1}{a^{2}}-\frac{1}{b^{2}}\).
Solution:
By sine rule,
\(\frac{\sin \mathrm{A}}{a}=\frac{\sin \mathrm{B}}{b}\)

(vii) \(\frac{b-c}{a}=\frac{\tan \frac{B}{2}-\tan \frac{C}{2}}{\tan \frac{B}{2}+\tan \frac{C}{2}}\)
Solution:
By sine rule,

Question 12.
In ∆ABC if a², b², c², are in A.P. then cot\(\frac{A}{2}\), cot\(\frac{B}{2}\), cot\(\frac{C}{2}\) are also in A.P.
Question is modified
In ∆ABC if a, b, c, are in A.P. then cot\(\frac{A}{2}\), cot\(\frac{B}{2}\), cot\(\frac{C}{2}\) are also in A.P.
Solution:
a, b, c, are in A.P.
∴ 2b = a + c …(1)

Question 13.
In ∆ABC if ∠C = 90º then prove that sin(A – B) = \(\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\)
Solution:
In ∆ABC, if ∠C = 90º
∴ c² = a² + b² …(1)
By sine rule,

Question 14.
In ∆ABC if \(\frac{\cos A}{a}=\frac{\cos B}{b}\), then show that it is an isosceles triangle.
Solution:
Given : \(\frac{\cos A}{a}=\frac{\cos B}{b}\) ….(1)
By sine rule,

∴ sin A cos B = cos A sinB
∴ sinA cosB – cosA sinB = 0
∴ sin (A – B) = 0 = sin0
∴ A – B = 0 ∴ A = B
∴ the triangle is an isosceles triangle.

Question 15.
In ∆ABC if sin²A + sin²B = sin²C then prove that the triangle is a right angled triangle.
Question is modified
In ∆ABC if sin²A + sin²B = sin²C then show that the triangle is a right angled triangle.
Solution:
By sine rule,
\(\frac{\sin \mathrm{A}}{a}\) = \(\frac{\sin \mathrm{B}}{b}\) = \(\frac{\sin \mathrm{C}}{c}\) = k
∴ sin A = ka, sinB = kb, sin C = kc
∴ sin²A + sin²B = sin²C
∴ k²a² + k²b² = k²c²
∴ a² + b² = c²
∴ ∆ABC is a rightangled triangle, rightangled at C.

Question 16.
In ∆ABC prove that a²(cos²B – cos²C) + b²(cos²C – cos²A) + c²(cos²A – cos²B) = 0.
Solution:
By sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\) = k
LHS = a²(cos²B – cos²C) + b²( cos²C – cos²A) + c²(cos²A – cos²B)
= k²sin²A [(1 – sin²B) – (1 – sin²C)] + k²sin²B [(1 – sin²C) – (1 – sin²A)] + k²sin²C[(1 – sin²A) – (1 – sin²B)]
= k²sin²A (sin²C – sin²B) + k²sin²B(sin²A – sin²C) + k²sin²C (sin²B – sin²A)
= k²(sin²A sin²C – sin²Asin²B + sin²A sin²B – sin²B sin²C + sin²B sin²C – sin²A sin²C)
= k²(0) = 0 = RHS.

Question 17.
With usual notations show that (c² – a² + b²) tan A = (a² – b² + c²) tan B = (b² – c² + a²) tan C.
Solution:
By sine rule,
\(\frac{a}{\sin A}\) = \(\frac{b}{\sin B}\) = \(\frac{c}{\sin C}\) = k
∴ a = fksinA, b = ksinB, c = ksinC


From (1), (2) and (3), we get
(c² – a² + b²) tan A = (a² – b² + c²) tan B
= (b² – c² + a²) tan C.

Question 18.
In ∆ABC, if a cos²\(\frac{C}{2}\) + c cos²\(\frac{A}{2}\) = \(\frac{3 b}{2}\), then prove that a , b ,c are in A.P.
Solution:
a cos²\(\frac{C}{2}\) + c cos²\(\frac{A}{2}\) = \(\frac{3 b}{2}\)

∴ a + c + b = 3b …[∵ a cos C + c cos A = b]
∴ a + c = 2b
Hence, a, b, c are in A.P.

Question 19.
Show that 2 sin^-1\(\left(\frac{3}{5}\right)\) = tan^-1\(\left(\frac{24}{7}\right)\).
Solution:
Let sin²\(\left(\frac{3}{5}\right)\) = x.



∴ tan^-1\(\left(\frac{24}{7}\right)\) = RHS

Question 20.
Show that tan^-1\(\left(\frac{1}{5}\right)\) + tan^-1\(\left(\frac{1}{7}\right)\) + tan^-1\(\left(\frac{1}{3}\right)\) + tan^-1\(\left(\frac{1}{8}\right)\) = \(\frac{\pi}{4}\).
Solution:

Question 21.
Prove that tan^-1\(\sqrt {x}\) = \(\frac{1}{2}\) cos^-1\(\left(\frac{1-x}{1+x}\right)\), if x ∈ [0, 1].
Solution:
Let tan^-1\(\sqrt {x}\) = y
∴ tan y = \(\sqrt {x}\) ∴ x = tan²y

Question 22.
Show that \(\frac{9 \pi}{8}-\frac{9}{4}\) sin^-1\(\frac{1}{3}\) = \(\frac{9}{4}\) sin^-1\(\frac{2 \sqrt{2}}{3}\).
Question is modified
Show that \(\frac{9 \pi}{8}-\frac{9}{4}\) sin^-1\(\left(\frac{1}{3}\right)\) = \(\frac{9}{4}\) sin^-1\(\left(\frac{2 \sqrt{2}}{3}\right)\).
Solution:
We have to show that

Question 23.
Show that

Solution:

Question 24.
If sin(sin^-1\(\frac{1}{5}\) + cos^-1x) = 1, then find the value of x.
Solution:
sin(sin^-1\(\frac{1}{5}\) = 1

Question 25.
If tan^-1\(\left(\frac{x-1}{x-2}\right)\) + tan^-1\(\left(\frac{x+1}{x+2}\right)\) = \(\frac{\pi}{4}\) then find the value of x.
Solution:
tan^-1\(\left(\frac{x-1}{x-2}\right)\) + tan^-1\(\left(\frac{x+1}{x+2}\right)\) = \(\frac{\pi}{4}\)

∴ x = ±\(\frac{1}{\sqrt{2}}\).

Question 26.
If 2 tan^-1(cos x ) = tan^-1(cosec x) then find the value of x.
Solution:
2 tan^-1(cos x ) = tan^-1(cosec x)

Question 27.
Solve: tan^-1\(\left(\frac{1-x}{1+x}\right)\) = \(\frac{1}{2}\)(tan^-1x), for x > 0.
Solution:
tan^-1\(\left(\frac{1-x}{1+x}\right)\) = \(\frac{1}{2}\)(tan^-1x)

Question 28.
If sin^-1(1 – x) – 2sin^-1x = \(\frac{\pi}{2}\), then find the value of x.
Solution:
sin^-1(1 – x) – 2sin^-1x = \(\frac{\pi}{2}\)

Question 29.
If tan^-12x + tan^-13x = \(\frac{\pi}{4}\), then find the value of x.
Question is modified
If tan^-12x + tan^-13x = \(\frac{\pi}{2}\), then find the value of x.
Solution:
tan^-12x + tan^-13x = \(\frac{\pi}{4}\)
∴ tan^-1\(\left(\frac{2 x+3 x}{1-2 x \times 3 x}\right)\) = tan\(\frac{\pi}{4}\), where 2x > 0, 3x > 0
∴ \(\frac{5 x}{1-6 x^{2}}\) = tan\(\frac{\pi}{4}\) = 1
∴ 5x = 1 – 6x²
∴ 6x² + 5x – 1 = 0
∴ 6x² + 6x – x – 1 = 0
∴ 6x(x +1) – 1(x + 1) = 0
∴ (x + 1)(6x – 1) = 0
∴ x = -1 or x = \(\frac{1}{6}\)
But x > 0 ∴ x ≠ -1
Hence, x = \(\frac{1}{6}\)

Question 30.
Show that tan^-1\(\frac{1}{2}\) – tan^-1\(\frac{1}{4}\) = tan^-1\(\frac{2}{9}\).
Solution:
LHS = tan^-1\(\frac{1}{2}\) – tan^-1\(\frac{1}{4}\)

Question 31.
Show that cot^-1\(\frac{1}{3}\) – tan^-1\(\frac{1}{3}\) = cot^-1\(\frac{3}{4}\).
Solution:
LHS = cot^-1\(\frac{1}{3}\) – tan^-1\(\frac{1}{3}\)

Question 32.
Show that tan^-1\(\frac{1}{2}\) = \(\frac{1}{3}\) tan^-1\(\frac{11}{2}\).
Solution:
We have to show that

Question 33.
Show that cos^-1\(\frac{\sqrt{3}}{2}\) + 2sin^-1\(\frac{\sqrt{3}}{2}\) = \(\frac{5 \pi}{6}\)
Solution:

Question 34.
Show that 2cot^-1\(\frac{3}{2}\) + sec^-1\(\frac{13}{12}\) = \(\frac{\pi}{2}\)
Solution:

Question 35.
Prove the following :
(i) cos^-1 x = tan^-1\(\frac{\sqrt{1-x^{2}}}{x}\), if x < 0.
Question is modified
cos^-1 x = tan^-1\(\left(\frac{\sqrt{1-x^{2}}}{x}\right)\), if x > 0.
Solution:

(ii) cos^-1 x = π + tan^-1\(\frac{\sqrt{1-x^{2}}}{x}\), if x < 0.
Solution:

Question 36.
If |x| < 1 , then prove that 2tan^-1 x = tan^-1\(\frac{2 x}{1-x^{2}}\) = sin^-1\(\frac{2 x}{1+x^{2}}\) = cos^-1\(\frac{1-x^{2}}{1+x^{2}}\)
Question is modified
If |x| < 1 , then prove that 2tan^-1 x = tan^-1\(\left(\frac{2 x}{1-x^{2}}\right)\) = sin^-1\(\left(\frac{2 x}{1+x^{2}}\right)\) = cos^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution:
Let tan^-1x = y
Then, x = tany

Question 37.
If x, y, z, are positive then prove that tan^-1\(\frac{x-y}{1+x y}\) + tan^-1\(\frac{y-z}{1+y z}\) + tan^-1\(\frac{z-x}{1+z x}\) = 0
Solution:

Question 38.
If tan^-1 x + tan^-1 y + tan^-1 z = \(\frac{\pi}{2}\) then, show that xy + yz + zx = 1
Solution:
tan^-1 x + tan^-1 y + tan^-1 z = \(\frac{\pi}{2}\)

∴ 1 – xy – yz – zx = 0
∴ xy + yz + zx = 1.

Question 39.
If cos^-1 x + cos^-1 y + cos^-1 z = π then show that x² + y² + z² + 2xyz = 1.
Solution:
0 ≤ cos^-1x ≤ π and
cos^-1x + cos^-1y+ cos^-1z = 3π
∴ cos^-1x = π, cos^-1y = π and cos^-1z = π
∴ x = y = z = cosπ = -1
∴ x² + y² + z² + 2xyz
= (-1)² + (-1)² + (-1)² + 2(-1)(-1)(-1)
= 1 + 1 + 1 – 2
= 3 – 2 = 1.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Ex 3.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Trigonometric Functions Ex 3.3

Question 1.
Find the principal values of the following :
(i) sin^-1\(\left(\frac{1}{2}\right)\)
Solution:
The principal value branch of sin^-1x is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Let sin^-1\(\left(\frac{1}{2}\right)\) = ∝, where \(\frac{-\pi}{2}\) ≤ ∝ ≤ \(\frac{\pi}{2}\)
∴ sin∝ = \(\frac{1}{2}\) = sin\(\frac{\pi}{6}\)
∴ ∝ = \(\frac{\pi}{6}\) …[∵ – \(\frac{\pi}{2}\) ≤ \(\frac{\pi}{6}\) ≤ \(\frac{\pi}{2}\)]
∴ the principal value of sin^-1\(\left(\frac{1}{2}\right)\) is \(\frac{\pi}{6}\).

(ii) cosec^-1(2)
Solution:
The principal value branch of cosec^-1x is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) – {0}.
Let cosec^-1(2) = ∝, where \(\frac{-\pi}{2}\) ≤ ∝ ≤ \(\frac{\pi}{2}\), ∝ ≠ 0
∴ cosec^-1 ∝ = 2 = cosec\(\frac{\pi}{6}\)
∴ ∝ = \(\frac{\pi}{6}\) …[∵ –\(\frac{\pi}{2}\) ≤ \(\frac{\pi}{6}\) ≤ \(\frac{\pi}{2}\)]
∴ the principal value of cosec^-1(2) is \(\frac{\pi}{6}\).

(iii) tan^-1(-1)
Solution:
The principal value branch of tan^-1x is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
Let tan^-1(-1) = ∝, where \(\frac{-\pi}{2}\) < ∝ < \(\frac{\pi}{2}\)
∴ tan∝ = -1 = -tan\(\frac{\pi}{4}\)
∴ tan∝ = tan\(\left(-\frac{\pi}{4}\right)\) …[∵ tan(-θ) = -tanθ]
∴ ∝ = –\(\frac{\pi}{4}\) …[∵ –\(\frac{\pi}{2}\) < \(\frac{-\pi}{4}\) < \(\frac{\pi}{2}\)]
∴ the principal value of tan^-1(-1) is –\(\frac{\pi}{4}\).

(iv) tan^-1(-\(\sqrt {3}\))
Solution:
The principal value branch of tan^-1x is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).
Let tan^-1(-\(\sqrt {3}\)) = ∝, where \(\frac{-\pi}{2}\) < ∝ < \(\frac{\pi}{2}\)
∴ tan∝ = –\(\sqrt {3}\) = -tan\(\frac{\pi}{3}\)
∴ tan∝ = tan\(\left(-\frac{\pi}{3}\right)\) …[∵ tan(-θ) = -tanθ]
∴ ∝ = –\(\frac{\pi}{3}\) …[∵ –\(\frac{\pi}{2}\) < \(\frac{-\pi}{3}\) < \(\frac{\pi}{2}\)]
∴ the principal value of tan^-1(-\(\sqrt {3}\)) is –\(\frac{\pi}{3}\).

(v) sin^-1 \(\left(\frac{1}{\sqrt{2}}\right)\)
Solution:
The principal value branch of sin^-1x is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Let sin^-1 \(\left(\frac{1}{\sqrt{2}}\right)\) = ∝, where \(\frac{-\pi}{2}\) < ∝ < \(\frac{\pi}{2}\)
∴ sin∝ = \(\left(\frac{1}{\sqrt{2}}\right)\) = sin\(\frac{\pi}{4}\)
∴ ∝ = \(\frac{\pi}{4}\) …[∵ –\(\frac{\pi}{2}\) ≤ \(\frac{\pi}{4}\) ≤ \(\frac{\pi}{2}\)]
∴ the principal value of sin^-1 \(\left(\frac{1}{\sqrt{2}}\right)\) is \(\frac{\pi}{4}\).

(vi) cos^-1\(\left(-\frac{1}{2}\right)\)
Solution:
The principal value branch of cos^-1x is (0, π).
Let cos^-1\(\left(-\frac{1}{2}\right)\) = ∝, where 0 ≤ ∝ ≤ π
∴ cos∝ = \(-\frac{1}{2}\) = -cos\(\frac{\pi}{3}\)
∴ cos∝ = cos\(\left(\pi-\frac{\pi}{3}\right)\) …[∵ cos(π – θ) = -cosθ)
∴ cos∝ = cos\(\frac{2 \pi}{3}\)
∴ ∝ = \(\frac{2 \pi}{3}\) …[∵ 0 ≤ \(\frac{2 \pi}{3}\) ≤ π]
∴ the principal value of cos^-1\(\left(-\frac{1}{2}\right)\) is \(\frac{2 \pi}{3}\).

Question 2.
Evaluate the following :
(i) tan^-1(1) + cos^-1\(\left(\frac{1}{2}\right)\) + sin^-1\(\left(\frac{1}{2}\right)\)
Solution:

(ii) cos^-1\(\left(\frac{1}{2}\right)\) + 2 sin^-1\(\left(\frac{1}{2}\right)\)
Solution:

(iii) tan^-1\(\sqrt {3}\) – sec^-1(-2)
Solution:

∴ tan^-1\(\sqrt {3}\) – sec^-1(-2)
= \(\frac{\pi}{3}-\frac{2 \pi}{3}\) …[By (1) and (2)]
= –\(\frac{\pi}{3}\).

(iv) cosec^-1( \(-\sqrt{2}\)) + cot^-1(\(\sqrt{3}\))
Solution:

Question 3.
Prove the following :
(i) sin^-1\(\left(\frac{1}{\sqrt{2}}\right)\) – 3sin^-1\(\left(\frac{\sqrt{3}}{2}\right)\) = –\(-\frac{3 \pi}{4}\)
Question is modified.
sin^-1\(\left(\frac{1}{\sqrt{2}}\right)\) – 3sin^-1\(\left(\frac{\sqrt{3}}{2}\right)\) = –\(\frac{3 \pi}{4}\)
Solution:

(ii) sin^-1\(\left(-\frac{1}{2}\right)\) + cos^-1\(\left(-\frac{\sqrt{3}}{2}\right)\) = cos^-1\(\left(-\frac{1}{2}\right)\)
Solution:

(iii) sin^-1\(\left(\frac{3}{5}\right)\) + cos^-1\(\left(\frac{12}{13}\right)\) = sin^-1\(\left(\frac{56}{65}\right)\)
Solution:

(iv) cos^-1\(\left(\frac{3}{5}\right)\) + cos^-1\(\left(\frac{4}{5}\right)\) = \(\frac{\pi}{2}\)
Solution:
Let cos^-1\(\left(\frac{3}{5}\right)\) = x
∴ cosx = \(\left(\frac{3}{5}\right)\), where 0 < x < \(\frac{\pi}{2}\) ∴ sinx > 0

(v) tan^-1\(\left(\frac{1}{2}\right)\) + tan^-1\(\left(\frac{1}{3}\right)\) = \(\frac{\pi}{4}\)
Solution:
LHS = tan^-1\(\left(\frac{1}{2}\right)\) + tan^-1\(\left(\frac{1}{3}\right)\)
= tan^-1\(\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}\right)\)
= tan^-1\(\left(\frac{3+2}{6-1}\right)\) = tan^-1(1)
= tan^-1\(\left(\tan \frac{\pi}{4}\right)\) = \(\frac{\pi}{4}\)
= RHS.

(vi) 2 tan^-1\(\left(\frac{1}{3}\right)\) = tan^-1\(\left(\frac{3}{4}\right)\)
Solution:

(vii) tan^-1\(\left[\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right]\) = \(\frac{\pi}{4}\) + θ if θ ∈ \(\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\)
Solution:

(viii) tan^-1\(\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\frac{\theta}{2}\), if θ ∈ (0, π)
Solution:

= \(\frac{\theta}{2}\) …[∵ tan^-1(tanθ) = θ]
= RHS.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Ex 3.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Trigonometric Functions Ex 3.2

Question 1.
Find the Cartesian co-ordinates of the point whose polar co-ordinates are:
(i) \(\left(\sqrt{2}, \frac{\pi}{4}\right)\)
Solution:
Here, r = \(\sqrt {2}\) and θ = \(\frac{\pi}{4}\)
Let the cartesian coordinates be (x, y)
Then, x = rcosθ = \(\sqrt {2}\)cos\(\frac{\pi}{4}\) = \(\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\) = 1
y = rsinθ = \(\sqrt {2}\)sin\(\frac{\pi}{4}\) = \(\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\) = 1
∴ the cartesian coordinates of the given point are (1, 1).

(ii) \(\left(4, \frac{\pi}{2}\right)\)
Solution:

(iii) \(\left(\frac{3}{4}, \frac{3 \pi}{4}\right)\)
Solution:
Here, r = \(\frac{3}{4}\) and θ = \(\frac{3 \pi}{4}\)
Let the cartesian coordinates be (x, y)

(iv) \(\left(\frac{1}{2}, \frac{7 \pi}{3}\right)\)
Solution:
Here, r = \(\frac{1}{2}\) and θ = \(\frac{7 \pi}{4}\)
Let the cartesian coordinates be (x, y)


∴ the cartesian coordinates of the given point are \(\left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)

Question 2.
Find the of the polar co-ordinates point whose Cartesian co-ordinates are.
(i) \((\sqrt{2}, \sqrt{2})\)
Solution:
Here x = \(\sqrt {2}\) and y = \(\sqrt {2}\)
∴ the point lies in the first quadrant.
Let the polar coordinates be (r, θ)
Then, r² = x² + y² = (\(\sqrt {2}\) )² + (\(\sqrt {2}\) )² = 2 + 2 = 4
∴ r = 2 … [∵ r > 0]
cos θ = \(\frac{x}{r}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\)
and sin θ = \(\frac{y}{r}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\)
∴ tan θ = 1
Since the point lies in the first quadrant and
0 ≤ θ ≤ 2π, tan θ = 1 = tan\(\frac{\pi}{4}\)
∴ θ = \(\frac{\pi}{4}\)
∴ the polar coordinates of the given point are \(\left(2, \frac{\pi}{4}\right)\).

(ii) \(\left(0, \frac{1}{2}\right)\)
Solution:
Here x = 0 and y = \(\frac{1}{2}\)
the point lies on the positive side of Y-axis. Let the polar coordinates be (r, θ)
Then, r² = x² + y² = (0)² + \(\left(\frac{1}{2}\right)^{2}=0+\frac{1}{4}=\frac{1}{4}\)
∴ r = \(\frac{1}{2}\) …[∵ r > 0]
cosθ = \(\frac{x}{r}=\frac{0}{(1 / 2)}\) = 0
and sin θ = \(\frac{y}{r}=\frac{(1 / 2)}{(1 / 2)}\) = 1
Since, the point lies on the positive side of Y-axis and 0 ≤ θ ≤ 2π
cosθ = 0 = cos\(\frac{\pi}{2}\) and sinθ = 1 = sin\(\frac{\pi}{2}\)
∴ θ = \(\frac{\pi}{2}\)
∴ the polar coordinates of the given point are \(\left(\frac{1}{2}, \frac{\pi}{2}\right)\).

(iii) \((1,-\sqrt{3})\)
Solution:
Here x = 1 and y = \(-\sqrt{3}\)
∴ the point lies in the fourth quadrant.
Let the polar coordinates be (r, θ).
Then, r² = x² + y² = (1)² + (\(-\sqrt {3}\) )² = 1 + 3 = 4
∴ r = 2 … [∵ r > 0]

∴ the polar coordinates of the given point are \(\left(2, \frac{5 \pi}{3}\right)\).

(iv) \(\left(\frac{3}{2}, \frac{3 \sqrt{3}}{2}\right)\)
Solution:

Question 3.
In ∆ABC, if ∠A = 45º, ∠B = 60º then find the ratio of its sides.
Solution:
By the sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\)
∴ \(\frac{a}{b}=\frac{\sin A}{\sin B}\) and \(\frac{b}{c}=\frac{\sin B}{\sin C}\)
∴ a : b : c = sinA : sinB : sinC
Given ∠A = 45° and ∠B = 60°
∵ ∠A + ∠B + ∠C = 180°
∴ 45° + 60° + ∠C = 180°
∴ ∠C = 180° – 105° = 75°

Question 4.
In ∆ABC, prove that sin \(\left(\frac{\mathbf{B}-\mathbf{C}}{2}\right)=\left(\frac{\boldsymbol{b}-\boldsymbol{c}}{a}\right)\) cos \(\frac{A}{2}\).
Solution:
By the sine rule,

Question 5.
With usual notations prove that 2 \(\left\{a \sin ^{2} \frac{C}{2}+c \sin ^{2} \frac{A}{2}\right\}\) = a – b + c.
Solution:

Question 6.
In ∆ABC, prove that a³sin(B – C) + b³sin(C – A) + c³sin(A – B) = 0
Solution:
By the sine rule,
\(\frac{a}{\sin A}\) = \(\frac{b}{\sin B}\) = \(\frac{c}{\sin C}\) = k
∴ a = k sin A, b = k sin B, c = k sin C
LHS = a³sin (B – C) + b³sin (C – A) + c³sin (A – B)
= a³(sin B cos C – cos B sin C) + b³(sinCcos A – cos C sin A) + c³(sinAcosB – cos A sin B)

= \(\frac{1}{2 k}\) [a²(a² + b² – c²) – a²(a² + c² – b²) + b²(b² + c² – a²) – b²(a² + b² – c²) + c²(c² + a² – b²) – c²(b² + c² – a²)]
= \(\frac{1}{2 k}\) [a⁴ + a²b² – a²c² – a⁴ – a²c² + a²b² + b⁴ + b²c² – a²b² – a²b² – b⁴ + b²c² + c⁴ + a²c² – b²c² – b²c² – c⁴ + a²c²]
= \(\frac{1}{2 k}\)(0) = 0 = RHS.

Question 7.
In ∆ABC, if cot A, cot B, cot C are in A.P. then show that a², b², c² are also in A.P
Solution:
By the sine rule,
\(\frac{\sin \mathrm{A}}{a}\) = \(\frac{\sin \mathrm{B}}{b}\) =\(\frac{\sin \mathrm{C}}{c}\) = k
∴ sin A = ka, sin B = kb, sin C = kc …(1)
Now, cot A, cotB, cotC are in A.P.
∴ cotC – cotB = cotB – cot A
∴ cotA + cotC = 2cotB

Question 8.
In ∆ABC, if a cos A = b cos B then prove that the triangle is right angled or an isosceles traingle.
Solution:
By the sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = k
a = k sin A and b = k sin B
∴ a cos A = b cos B gives
k sin A cos A = k sin B cos B
∴ 2 sin A cos A = 2 sin B cos B
∴ sin 2A = sin 2B ∴ sin 2A – sin 2B = 0
∴ 2 cos (A + B)∙sin (A -B) = 0
∴ 2cos (π – C)∙sin(A – B) = 0 … [∵ A + B + C = π]
∴ -2 cos C∙sin (A – B) = 0
∴ cos C = 0 OR sin(A -B) = 0
∴ C = 90° OR A – B = 0
∴ C = 90° OR A = B
∴ the triangle is either rightangled or an isosceles triangle.

Question 9.
With usual notations prove that 2(bc cos A + ac cos B + ab cos C) = a² + b² + c².
Solution:
LHS = 2 (bc cos A + ac cos B + ab cos C)
= 2bc cos A + 2ac cos B + 2ab cos C
= 2bc \(\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)\) + 2ac\(\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)\) + 2ab\(\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)\) …(By cosine rule]
= b² + c² – a² + c² + a² – b² + a² + b² – c² = a² + b² + c² = RHS.

Question 10.
In △ABC, if a = 18, b = 24, c = 30 then find the values of
(i) cos A
Solution:
Given : a = 18, b = 24 and c = 30
∴ 2s = a + b + c = 18 + 24 + 30 = 72 ∴ s = 36

(ii) sin\(\frac{A}{2}\)
Solution:

(iii) cos\(\frac{A}{2}\)
Solution:

(iv) tan\(\frac{A}{2}\)
Solution:

(v) A(△ABC)
Solution:

(iv) sin A.
Solution:

Question 11.
In △ABC prove that (b + c – a) tan \(\frac{A}{2}\) = (c + a – b) tan\(\frac{B}{2}\) = (a + b – c) tan\(\frac{C}{2}\).
Solution:
(b + c – a) tan \(\frac{A}{2}\)

Question 12.
In △ABC prove that sin \(\frac{A}{2}\)∙sin \(\frac{A}{2}\)∙sin \(\frac{A}{2}\) = \(\frac{[A(\triangle A B C)]^{2}}{a b c s}\)
Solution:
LHS = sin \(\frac{A}{2}\)∙sin \(\frac{B}{2}\)∙sin \(\frac{C}{2}\)

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Ex 3.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Trigonometric Functions Ex 3.1

Question 1.
Find the principal solutions of the following equations :
(i) cos θ= \(\frac{1}{2}\)
Solution:
We know that, cos\(\frac{\pi}{3}\) = \(\frac{1}{2}\) and cos (2π – θ) = cos θ
∴ cos\(\frac{\pi}{3}\) = cos(2π – \(\frac{\pi}{3}\)) = cos\(\frac{5 \pi}{3}\)
∴ cos\(\frac{\pi}{3}\) = cos\(\frac{5 \pi}{3}\) = \(\frac{1}{2}\), where
0 < \(\frac{\pi}{3}\) < 2π and 0 < \(\frac{5 \pi}{3}\) < 2π
∴ cos θ = \(\frac{1}{2}\) gives cos θ = cos\(\frac{\pi}{3}\) = cos\(\frac{5 \pi}{3}\)
∴ θ = \(\frac{\pi}{3}\) and θ = \(\frac{5 \pi}{3}\)
Hence, the required principal solutions are
θ = \(\frac{\pi}{3}\) and θ = \(\frac{5 \pi}{3}\)

(ii) sec θ = \(\frac{2}{\sqrt{3}}\)
Solution:

(iii) cot θ = \(\sqrt {3}\)
Solution:
The given equation is cot θ = \(\sqrt {3}\) which is same as tan θ = \(\frac{1}{\sqrt{3}}\).
We know that,

Hence, the required principal solution are
θ = \(\frac{\pi}{6}\) and θ = \(\frac{7 \pi}{6}\).

(iv) cot θ = 0.
Solution:

Question 2.
Find the principal solutions of the following equations:
(i) sinθ = \(-\frac{1}{2}\)
Solution:
We know that,
sin\(\frac{\pi}{6}\) = \(\frac{1}{2}\) and sin (π + θ) = -sinθ,
sin(2π – θ) = -sinθ

Hence, the required principal solutions are
θ = \(\frac{7\pi}{6}\) and θ = \(\frac{11 \pi}{6}\).

(ii) tanθ = -1
Solution:
We know that,
tan\(\frac{\pi}{4}\) = 1 and tan(π – θ) = -tanθ,
tan(2π – θ) = -tanθ

Hence, the required principal solutions are
θ = \(\frac{3\pi}{4}\) and θ = \(\frac{7 \pi}{4}\).

(iii) \(\sqrt {3}\) cosecθ + 2 = 0.
Solution:

Question 3.
Find the general solutions of the following equations :
(i) sinθ = \(\frac{1}{2}\)
Solution:
(i) The general solution of sin θ = sin ∝ is
θ = nπ + (-1 )ⁿ∝, n ∈ Z
Now, sinθ = \(\frac{1}{2}\) = sin\(\frac{\pi}{6}\) …[∵ sin\(\frac{\pi}{6}\) = \(\frac{1}{2}\)]
∴ the required general solution is
θ = nπ + (-1)ⁿ\(\frac{\pi}{6}\), n ∈ Z.

(ii) cosθ = \(\frac{\sqrt{3}}{2}\)
Solution:
The general solution of cos θ = cos ∝ is
θ = 2nπ ± ∝, n ∈ Z
Now, cosθ = \(\frac{\sqrt{3}}{2}\) = cos\(\frac{\pi}{6}\) …[∵ cos\(\frac{\pi}{6}\) = \(\frac{\sqrt{3}}{2}\)]
∴ the required general solution is
θ = 2nπ ± \(\frac{\pi}{6}\), n ∈ Z.

(iii) tanθ = \(\frac{1}{\sqrt{3}}\)
Solution:
The general solution of tan θ = tan ∝ is
θ = nπ + ∝, n ∈ Z
Now, tan θ = \(\frac{1}{\sqrt{3}}\) = tan\(\frac{\pi}{6}\) …[tan\(\frac{\pi}{6}\) = \(\frac{1}{\sqrt{3}}\)]
∴ the required general solution is
θ = nπ + \(\frac{\pi}{6}\) , n ∈ Z.

(iv) cotθ = 0.
Solution:
The general solution of tan θ = tan ∝ is
θ = nπ + ∝, n ∈ Z
Now, cot θ = 0 ∴ tan θ does not exist
∴ tanθ = tan\(\frac{\pi}{2}\) [∵ tan\(\frac{\pi}{2}\) does not exist]
∴ the required general solution is
θ = nπ + \(\frac{\pi}{2}\), n ∈ Z.

Question 4.
Find the general solutions of the following equations:
(i) secθ = \(\sqrt {2}\)
Solution:
The general solution of cos θ = cos ∝ is
θ = nπ ± ∝, n ∈ Z.
Now, secθ = \(\sqrt {2}\) ∴ cosθ = \(\frac{1}{\sqrt{2}}\)
∴ cosθ = cos\(\frac{\pi}{4}\) ….[cos\(\frac{\pi}{4}\) = \(\frac{1}{\sqrt{2}}\)]
∴ the required general solution is
θ = 2nπ ± \(\frac{\pi}{4}\), n ∈ Z.

(ii) cosecθ = –\(\sqrt {2}\)
Solution:
The general solution of sinθ = sin∝ is

(iii) tanθ = -1
Solution:
The general solution of tanθ = tan∝ is

Question 5.
Find the general solutions of the following equations :
(i) sin 2θ = \(\frac{1}{2}\)
Solution:
The general solution of sin θ = sin ∝ is
θ = nπ + (-1)ⁿ∝, n ∈ Z

(ii) tan \(\frac{2 \theta}{3}\) = \(\sqrt {3}\)
Solution:
The general solution of tan θ = tan ∝ is

(iii) cot 4θ = -1
Solution:
The general solution of tan θ = tan ∝ is

Question 6.
Find the general solutions of the following equations :
(i) 4 cos²θ = 3
Solution:
The general solution of cos²θ = cos² ∝ is
θ = nπ ± ∝, n ∈ Z
Now, 4 cos²θ = 3

(ii) 4 sin²θ = 1
Solution:
The general solution of sin²θ = sin² ∝ is
θ = nπ ± ∝, n ∈ Z
Now, 4 sin²θ = 3

(iii) cos 4θ = cos 2θ
Solution:
The general solution of cos θ = cos ∝ is
θ = 2nπ ± ∝, n ∈ Z
∴ the general solution of cos 4θ = cos 2θ is given by
4θ = 2nπ ± 2θ, n ∈ Z
Taking positive sign, we get
4θ = 2nπ + 2θ, n ∈ Z
∴ 2θ = 2nπ, n ∈ Z
∴ θ = nπ, n ∈ Z
Taking negative sign, we get
4θ = 2nπ – 2θ, n ∈ Z
∴ 6θ = 2nπ, n ∈ Z
∴ θ = \(\frac{n \pi}{3}\), n ∈ Z
Hence, the required general solution is
θ = \(\frac{n \pi}{3}\), n ∈ Z or ∴ θ = nπ, n ∈ Z.
Alternative Method:
cos 4θ = cos 2θ
∴ cos4θ – cos 20 = 0
∴ -2sin\(\left(\frac{4 \theta+2 \theta}{2}\right)\)∙sin\(\left(\frac{4 \theta-2 \theta}{2}\right)\) = 0
∴ sin3θ∙sinθ = 0
∴ either sin3θ = 0 or sin θ = 0
The general solution of sin θ = 0 is
θ = nπ, n ∈ Z.
∴ the required general solution is given by
3θ = nπ, n ∈ Z or θ = nπ, n ∈ Z
i.e. θ = \(\frac{n \pi}{3}\), n ∈ Z or θ = nπ, n ∈ Z.

Question 7.
Find the general solutions of the following equations :
(i) sinθ = tanθ
Solution:
sin θ = tan θ
∴ sin θ = \(\frac{\sin \theta}{\cos \theta}\)
∴ sin θ cos θ = sin θ
∴ sin θ cos θ – sinθ = 0
∴ sin θ (cos θ – 1) = θ
∴ either sinθ = 0 or cosθ – 1 = 0
∴ either sin θ = 0 or cos θ = 1
∴ either sinθ = 0 or cosθ = cosθ …[∵ cos0 = 1]
The general solution of sinθ = 0 is θ = nπ, n ∈ Z and cos θ = cos ∝ is θ = 2nπ ± ∝, where n ∈ Z.
∴ the required general solution is given by
θ = nπ, n ∈ Z or θ = 2nπ ± 0, n ∈ Z
∴ θ = nπ, n ∈ Z or θ = 2nπ, n ∈ Z.

(ii) tan³θ = 3tanθ
Solution:
tan³θ = 3tanθ
∴ tan³θ – 3tanθ = 0
∴ tan θ (tan²θ – 3) = 0
∴ either tan θ = 0 or tan²θ – 3 = 0
∴ either tanθ = 0 or tan²θ = 3
∴ either tan θ = 0 or tan²θ = (\(\sqrt {3}\) )³
∴ either tan θ = 0 or tan²θ = (tan\(\frac{\pi}{3}\))³ …[tan\(\frac{\pi}{3}\) = \(\sqrt {3}\)]
∴ either tanθ = 0 or tan²θ = tan²\(\frac{\pi}{3}\)
The general solution of
tanθ = 0 is θ = nπ, n ∈ Z and
tan²θ = tan²∝ is θ = nπ ± ∝, n ∈ Z.
∴ the required general solution is given by
θ = nπ, n ∈ Z or θ = nπ ± \(\frac{\pi}{3}\), n ∈ Z.

(iii) cosθ + sinθ = 1.
Solution:
cosθ + sinθ = 1

Question 8.
Which of the following equations have solutions ?
(i) cos 2θ = -1
Solution:
cos 2θ = -1
Since -1 ≤ cos θ ≤ 1 for any θ,
cos 2θ = -1 has solution.

(ii) cos²θ = -1
Solution:
cos²θ = -1
This is not possible because cos²θ ≥ 0 for any θ.
∴ cos²θ = -1 does not have any solution.

(iii) 2 sinθ = 3
Solution:
2 sin θ = 3 ∴ sin θ = \(\frac{3}{2}\)
This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 2 sin θ = 3 does not have any solution.

(iv) 3 tanθ = 5
Solution:
3tanθ = 5 ∴ tanθ = \(\frac{5}{3}\)
This is possible because tan θ is any real number.
∴ 3tanθ = 5 has solution.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Miscellaneous Exercise 2B Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B

I. Choose the correct answer from the given alternatives in each of the following questions:

Question 1.
If A = \(\left(\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right)\), adj = \(\left(\begin{array}{ll}
4 & a \\
-3 & b
\end{array}\right)\) then the values of a and b are,
(a) a = – 2, b = 1
(b) a = 2, b = 4
(c) a = 2, b = –1
(d) a = 1, b = –2
Solution:
(a) a = – 2, b = 1

Question 2.
The inverse of \(\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right)\) is

Solution:
\(\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right)\)

Question 3.
If A = \(\left(\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right)\) and A(adj A) = k 1, then the value of k is
(a) 1
(b) -1
(c) 0
(d) -3
Solution:
(d) -3 [Hint : A(adj A) = |A| ∙ I]

Question 4.
If A = \(\left(\begin{array}{ll}
2 & -4 \\
3 & 1
\end{array}\right)\), then the adjoint of matrix A is

Solution:\(\left(\begin{array}{ll}
1 & 4 \\
-3 & 2
\end{array}\right)\)

Question 5.
If A = \(\left(\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right)\) and A(adj A) = kI, then the value of k is
(a) 2
(b) -2
(c) 10
(d) -10
Solution:
(b) -2

Question 6.
If A = \(\left(\begin{array}{rr}
\lambda & 1 \\
-1 & -\lambda
\end{array}\right)\), then A^-1 does not exist if λ = ………..
(a) 0
(b) ± 1
(c) 2
(d) 3
Solution:
(b) ± 1

Question 7.
If A = \(\left[\begin{array}{ll}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) then A^-1 = ….

Solution:
\(\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\)

Question 8.
If F (∝) = \(\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\) where ∝ ∈ R then [F(∝)]^-1 is =
(a) F(-∝)
(b) F(∝^-1)
(c) F(2∝)
(d) None of these
Solution:
(a) F(-∝)

Question 9.
The inverse of A = \(\left[\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right]\)
(a) I
(b) A
(c) A’
(d) -I
Solution:
(b) A

Question 10.
The inverse of a symmetric matrix is
(a) Symmetric
(b) Non-symmetric
(c) Null matrix
(d) Diagonal matrix
Solution:
(a) Symmetric

Question 11.
For a 2 × 2 matrix A, if A(adjA) = \(\left(\begin{array}{ll}
10 & 0 \\
0 & 10
\end{array}\right)\) then determinant A equals
(a) 20
(b) 10
(c) 30
(d) 40
Solution:
(b) 10

Question 12.
If A² = \(-\frac{1}{2}\left[\begin{array}{cc}
1 & -4 \\
-1 & 2
\end{array}\right]\) then A =

Solution:
\(-\frac{1}{2}\left[\begin{array}{cc}
2 & 4 \\
1 & 1
\end{array}\right]\)

II. Solve the following equations by the methods of inversion.

(i) 2x – y = -2 , 3x + 4y = 5
Solution:
The given equations can be written in the matrix form as :



By equality of matrices,
x = \(-\frac{5}{11}\), y = \(\frac{12}{11}\) is the required solution.

(ii) x + y + z = 1, 2x + 3y + 2z = 2 and ax + ay + 2az = 4, a ≠ 0.
Solution:
The given equations can be written in the matrix form as :

= 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a)
= 4a – 2a – a = a ≠ 0 ∴ A^-1 exists.
Consider AA^-1 = I

(iii) 5x – y +4z = 5, 2x + 3y + 5z = 2 and 5x – 2y + 6z = -1
Solution:
The given equations can be written in the matrix form as :

= 5(18 + 10) + 1 (12 – 25) + 4( -4 – 15)
= 140 – 13 – 76 = 51 #0
∴ A^-1 exists.
Now, we have to find the cofactor matrix


Now, premultiply AX = B by A^-1, we get,
A^-1(AX) = A^-1B
∴ (A^-1A)X = A^-1B
∴ IX = A^-1B
∴ X = \(\frac{1}{51}\left[\begin{array}{rrr}
28 & -2 & -17 \\
13 & 10 & -17 \\
-19 & 5 & 17
\end{array}\right]\left[\begin{array}{r}
5 \\
2 \\
-1
\end{array}\right]\)

By equality of matrices,
x = 3, y = 2, z = -2 is the required solution.

(iv) 2x + 3y = -5, 3x + y = 3
Solution:

(v) x + y + z = -1, y + z = 2 and x + y – z = 3
Solution:
The given equations can be written in the matrix form as :

= 1(-1 – 1) – 1 (0 – 1) + 1(0 – 1)
= -2 + 1 – 1 = -2 ≠ 0 ∴ A^-1 exists.
Consider AA^-1 = I


Now, premultiply AX = B by A^-1, we get,
A^-1(AX) = A^-1B
∴ (A^-1A)X = A^-1B
∴ IX = A^-1B

∴ by equality of the matrices, x= -3, y = 4, z = -2 is the required solution.

Question 2.
Express the following equation in matrix from and solve them by the method of reduction.
(i) x – y + z = 1, 2x – y = 1, 3x + 3y – 4z = 2
Solution:
The given equations can be written in the matrix form as :


By equality of matrices,
x – y + z = 1 ……(1)
y – 2z = -1 …..(2)
5z = 5 ….(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
y – 2 = -1 ∴ y = 1
Substituting y = 1, z = 1 in (1), we get,
x – 1 + 1 = 1
∴ x = 1
Hence, x = 1, y = 1, z = 1 is the required solution.

(ii) x + y = 1, y + z = \(\frac{5}{3}\), z + x = \(\frac{4}{3}\).
Solution:
The given equations can be written in the matrix form as :

By equality of matrices,
x + y = 1 ……(1)
y + z = \(\frac{5}{3}\) …(2)
2z = 2 ……..(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
y + 1 = \(\frac{5}{3}\) ∴ y = \(\frac{2}{3}\)
Substituting y = \(\frac{2}{3}\) in (1), we get,
x + \(\frac{2}{3}\) = 1 ∴ x = \(\frac{1}{3}\)
Hence, x = \(\frac{1}{3}\), y = \(\frac{2}{3}\), z = 1 is the required solution.

(iii) 2x – y + z = 1, x + 2y + 3z = 8 and 3x + y – 4z = 1
Solution:
The given equations can be written in the matrix form as :

∴ \(\left[\begin{array}{r}
x+2 y+3 z \\
0-5 y-5 z \\
0+0-8 z
\end{array}\right]\) = \(\left[\begin{array}{r}
8 \\
-15 \\
-8
\end{array}\right]\)
By equality of matrices,
x + 2y + 3z = 8 …..(1)
-5y – 5z = -15 ….(2)
-8z = -8 …..(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
-5y – 5 = -15
-5y = -10
∴ y = 2
Substituting y = 2, z = 1 in (1), we get,
x + 4 + 3 = 8 ∴ x = 1
Hence, x = 1, y = 2, z = 1 is the required solution.

(iv) x + y + z = 6, 3x – y + 3z =10 and 5x + 5y – 4z = 3.
Solution:

(v) x + 2y + z = 8, 2x + 3y – z =11 and 3x – y – 2z = 5
Solution:
The given equations can be written in the matrix form as :

By equality of matrices,
x + 2y + z = 8 … (1)
-y – 3z = -5 … (2)
16z = 16 … (3)
From (3), z = 1
Substituting z = 1 in (2), we get,
-y – 3 = -5, ∴ y = 2
Substituting y = 2, z = 1 in (1), we get,
x + 4 + 1 = 8 ∴ x = 3
Hence, x = 3, y = 2, z = 1 is the required solution.

(vi) x + 3y + 2z = 6, 3x – 2y + 5z =5 and 2x – 3y + 6z = 7.
Solution:
The given equations can be written in the matrix form as :


By equality of matrices,
x + 3y + 2z = 6 …(1)
y + \(\frac{3}{2}\) z = 4 …(2)
\(\frac{31}{2}\)z = 31 …..(3)
From (3), z = 2
Substituting z = 2 in (2), we get,
y + \(\frac{3}{2}\)z = 4
y + \(\frac{3}{2}\)(2) = 4
y + 3 = 4
y = 1
Substituting y = 1, z = 2 in (2), we get,
x + 3y + 2z = 6
x + 3(1) + 2(2) = 6
x + 3 + 4 = 6
x = -1
Hence, x = -1, y = 1, z = 2 is the required solution.

Question 3.
The sum of three numbers is 6. If we multiply third number by 3 and add it to the second number we get 11. By adding first and the third numbers we get a number which is double the second number. Use this information and find a system of linear equations. Find the three numbers using matrices.
Solution:
Let the three numbers be x, y and z. According to the given conditions,
x + y + z = 6.
3z + y = 11, i.e., y + 3z = 11 and x + z = 2y,
i.e., x – 2y + z = 0
Hence, the system of the linear equations is
x + y + z = 6
y + 3z = 11
x – 2y + z = 0
These equations can be written in the matrix form as :

By equality of matrices,
x + y + z = 6 …(1)
y + 3z = 11 …(2)
-3y = -6 …(3)
From (3), y = 2
Substituting y = 2 in (2), we get,
2 + 3z = 11
∴ 3z = 9 ∴ z = 3
Put y = 2, z — 3 in (1), we get,
x + 2 + 3 = 6 ∴ x = 1
∴ x = 1, y = 2, z = 3
Hence, the required numbers are 1, 2 and 3.

Question 4.
The cost of 4 pencils, 3 pens and 2 books is ₹ 150. The cost of 1 pencil, 2 pens and 3 books is ₹ 125. The cos of 6 pencils, 2 pens and 3 books is ₹ 175. Fild the cost of each item by using Matrices.
Solution:
Let the cost of 1 pencil, 1 pen and 1 book be ₹x, ₹ y, ₹ z respectively.
According to the given conditions,
4x + 3y + 2z = 150
x + 2y + 3z = 125
6x + 2y + 3z = 175
The equations can be written in matrix form as :

By equality of matrices,
x + 2y + 3z = 125 …(1)
-5y – 10z = -350 …(2)
5z = 125 …(3)
From (3), z = 25
Substituting z = 25 in (2), we get
-5y – 10(25) = -350
∴ -5y = -350 + 250 = -100
∴ y = 20
Substituting y = 20, z = 25 in (1), we get
x + 2(20) + 3(25) = 125
∴ x = 125 – 40 – 75 = 10
∴ x = 10, y = 20, z = 25
Hence, the cost of 1 pencil is ₹ 10, 1 pen is ₹ 20 and 1 book is ₹ 25.

Question 5.
The sum of three numbers is 6. Thrice the third number when added to the first number, gives 7. On adding three times first number to the sum of second and third number, we get 12. Find the three numbers by using Matrices.
Solution:
Let the numbers be x, y and z.
According to the given conditions,
x + y + z = 6
3z + x = 7, i.e., x + 3z = 7
and 3x + y + z = 12
Hence, the system of linear equations is
x + y + z = 6
x + 3z = 7
3x + y + z = 12
These equations can be written in matrix form as :

By equality of matrices,
x + y + z = 6 …(1)
-y + 2z = 1 …(2)
-3y = -5 …(3)
From (3), y = \(\frac{5}{3}\)
Substituting y = \(\frac{5}{3}\) in (2), we get,
–\(\frac{5}{3}\) + 2z = 1
∴ 2z = 1 + \(\frac{5}{3}\) = \(\frac{8}{3}\)
∴ z = \(\frac{4}{3}\)
Substituting y =\(\frac{5}{3}\), z = \(\frac{5}{3}\) in (1), we get,
x + \(\frac{5}{3}+\frac{4}{3}\) = 6
∴ x = 3
∴ x = 3, y = \(\frac{5}{3}\), z = \(\frac{4}{3}\)
Hence, the required numbers are 3, \(\frac{5}{3}\) and \(\frac{4}{3}\).

Question 6.
The sum of three numbers is 2. If twice the second number is added to the sum of first and third number, we get 1 adding five times the first number to the sum of second and third we get 6. Find the three numbers by using matrices.
Solution:
Let the three numbers be x, y and z.
According to the question,
x + y + 2
x + 2y + z = 1
5x + y + z = 6
The given system of equations can be written in matrix form as follows:

Question 7.
An amount of ₹ 5000 is invested in three types of investments, at interest rates 6%, 7%, 8% per annum respectively. The total annual income from this investment is ₹ 350. If the total annual income from the first two investments is ₹ 70 more than the income from the third, find the amount of each investment using the matrix method.
Solution:
Let the amounts in three investments by ₹ x, ₹ y, and ₹ z respectively.
Then x + y + z = 5000
Since the rate of interest in these investments are 6%, 7% and 8% respectively, the annual income of the three investments are \(\frac{6 x}{100}\), \(\frac{7 y}{100}\) and \(\frac{8 z}{100}\) respectively.
According to the given conditions,
\(\frac{6 x}{100}+\frac{7 y}{100}+\frac{8 z}{100}\) = 350
i.e. 6x + 7y + 8z = 35000
Also, \(\frac{6 x}{100}+\frac{7 y}{100}\) = \(\frac{8 z}{100}\) + 70
i.e. 6x + 7y – 8z = 7000
Hence, the system of linear equation is
x + y + z = 5000
6x + 7y + 8z = 35000
6x + 7y – 8z = 7000
These equations can be written in matrix form as :


By equality of matrices,
x + y + z = 5000 …(1)
y + 2z = 5000 …(2)
-16z = -28000 ….(3)
From (3), z = 1750
Substituting z = 1750 in (2), we get,
y + 2(1750) = 5000
∴ y = 5000 – 3500 = 1500
Substituting y = 1500, z = 1750 in (1), we get,
x + 1500 + 1750 = 5000
∴ x = 5000 – 3250 = 1750
∴ x = 1750, y = 1500, z = 1750
Hence, the amounts of the three investments are ₹ 1750, ₹ 1500 and ₹ 1750 respectively.

Question 8.
The sum of the costs of one ook each of Mathematics, Physics and Chemistry is ₹ 210. Total cost of a mathematics book, 2 physics books, and a chemistry book is ₹ 240 Also the total cost of a Mathematics book, 3 physics book and chemistry books is Rs. 300/-. Find the cost of each book, using Matrices.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Miscellaneous Exercise 2A Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2A

Question 1.
If A = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right]\) then reduce it to I₃ by using column transformations.
Solution:
|A| = \(\left|\begin{array}{lll}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right|\)
= 1(1 – 0) – 0 + 0 = 1 ≠ 0
∴ A is a non-singular matrix.
Hence, the required transformation is possible.
Now, A = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right]\)
By C₁ – 2C₂, we get, A ~ \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
-3 & 3 & 1
\end{array}\right]\)
By C₁ + 3C₃ and C₂ – 3C₃, we get,
A ~ \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = I₃.

Question 2.
If A = \(\left[\begin{array}{lll}
2 & 1 & 3 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{array}\right]\), then reduce it to I₃ by using row transformations.
Solution:
|A| = \(\left|\begin{array}{lll}
2 & 1 & 3 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{array}\right|\)
= 2 (0 – 1) – 1(1 – 1) + 3 (1 – 0)
= -2 – 0 + 3 = 1 ≠ 0
∴ A is a non-singular matrix.
Hence, the required transformation is possible.
Now, A = \(\left[\begin{array}{lll}
2 & 1 & 3 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{array}\right]\)
By R₁ – R₂, we get,

By R₁ – R₃ and By R₂ – R₃, we get
A ~ \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = I₃.

Question 3.
Check whether the following matrices are invertible or not:
(i) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\) = 1 – 0 = 1 ≠ 0.
∴ A is a non-singular matrix.
Hence, A^-1 exists.

(ii) \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right|\) = 1 – 1 = 0.
∴ A is a singular matrix.
Hence, A^-1 does not exist.

(iii) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 3
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{ll}
1 & 2 \\
3 & 3
\end{array}\right|\) = 3 – 6 = -3 ≠ 0.
∴ A is a non-singular matrix.
Hence, A^-1 exist.

(iv) \(\left[\begin{array}{ll}
2 & 3 \\
10 & 15
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
2 & 3 \\
10 & 15
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{ll}
2 & 3 \\
10 & 15
\end{array}\right|\) = 30 – 30 = 0.
∴ A is a singular matrix.
Hence, A^-1 does not exist.

(v) \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{cc}
\sec \theta & \tan \theta \\
\tan \theta & \sec \theta
\end{array}\right|\)
= sec²θ – tan²θ = 1 ≠ 0.
∴ A is a non-singular matrix.
Hence, A^-1 exist.

(vii) \(\left[\begin{array}{lll}
3 & 4 & 3 \\
1 & 1 & 0 \\
1 & 4 & 5
\end{array}\right]\)
Solution:
let A = \(\left[\begin{array}{lll}
3 & 4 & 3 \\
1 & 1 & 0 \\
1 & 4 & 5
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{lll}
3 & 4 & 3 \\
1 & 1 & 0 \\
1 & 4 & 5
\end{array}\right|\)
= 3(5 – 0) – 4(5 – 0) + 3(4 – 1)
= 15 – 20 + 9 = 4 ≠ 0
∴ A is a non-singular matrix.
Hence, A^-1 exist.

(viii) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & -1 & 3 \\
1 & 2 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & -1 & 3 \\
1 & 2 & 3
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
2 & -1 & 3 \\
1 & 2 & 3
\end{array}\right|\)
= 1 (-3 -6) – 2 (6 – 3) + 3 (4 + 1)
= -9 – 6 + 15 = 0
∴ A is a singular matrix.
Hence, A^-1 does not exist.

(ix) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 4 & 5 \\
4 & 6 & 8
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 4 & 5 \\
4 & 6 & 8
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
3 & 4 & 5 \\
4 & 6 & 8
\end{array}\right|\)
= 1(32 – 30) – 2(24 – 20) + 3(18 – 16)
= 2 – 8 + 6 = 0
∴ A is a singular matrix.
Hence, A^-1 does not exist.

Question 4.
Find AB, if A = \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
1 & -2 & -3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
1 & -1 \\
1 & 2 \\
1 & -2
\end{array}\right]\) Examine whether AB has inverse or not.
Solution:

∴ A is a non-singular matrix.
Hence, (AB)^-1 exist.

Question 5.
If A = \(\left[\begin{array}{lll}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z
\end{array}\right]\) is a nonsingular matrix then find A^-1 by elementary row transformations.
Hence, find the inverse of \(\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right]\)
Solution:
Since A is a non-singular matrix, then find A^-1 by using elementary row transformations.
We write AA^-1 = I

Comparing \(\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right]\) with \(\left[\begin{array}{lll}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z
\end{array}\right]\),
we get, x = 2, y = 1, z = -1
∴ \(\frac{1}{x}\) = \(\frac{1}{2}\), \(\frac{1}{y}\) = \(\frac{1}{1}\) = 1, \(\frac{1}{z}\) = \(\frac{1}{-1}\) = -1
\(\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right]\) is \(\left(\begin{array}{rrr}
\frac{1}{2} & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right)\).

Question 6.
if A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\) and X is a 2 × 2 matrix such that AX = I , then find X.
Solution:
We will reduce the matrix A to the identity matrix by using row transformations. During this pro¬cess, I will be converted to the matrix X.
We have AX = I.

Question 7.
Find the inverse of each of the following matrices (if they exist).
(i) \(\left[\begin{array}{ll}
1 & -1 \\
2 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & -1 \\
2 & 3
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
1 & -1 \\
2 & 3
\end{array}\right|\) = 3 + 2 = 5 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(ii) \(\left[\begin{array}{ll}
2 & 1 \\
1 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & -1
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
2 & 1 \\
1 & -1
\end{array}\right|\) = -2 – 1 = -3 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(iii) \(\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right|\) = 7 – 6 = 1 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(iv) \(\left[\begin{array}{ll}
2 & -3 \\
5 & 7
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
2 & -3 \\
5 & 7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
2 & -3 \\
5 & 7
\end{array}\right|\) = 14 + 15 = 29 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(v) \(\left[\begin{array}{ll}
2 & 1 \\
7 & 4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
2 & 1 \\
7 & 4
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
2 & 1 \\
7 & 4
\end{array}\right|\) = 8 – 7 = 1 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(vi) \(\left[\begin{array}{ll}
3 & -10 \\
2 & -7
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
3 & -10 \\
2 & -7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
3 & -10 \\
2 & -7
\end{array}\right|\) = -21 + 20 = -1 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(vii) \(\left[\begin{array}{lll}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
\end{array}\right|\)
= 2(4 + 6) +3(4 – 9) + 3(-4 – 6)
= 20 – 15 – 30 = -25 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(viii) \(\left[\begin{array}{lll}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right|\)
= 1(0 + 25) + 3(0 + 10) + 2(-15 – 0)
= 25 + 30 -30
= 25 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(ix) \(\left[\begin{array}{lll}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
Solution:
Let A =\(\left[\begin{array}{lll}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right|\)
= 2(3 – 0) – 0 – 1(5 – 0)
= 6 – 0 – 5 = 1 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I


∴ A^-1 = \(\left[\begin{array}{lll}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\)

(x) \(\left[\begin{array}{lll}
1 & 2 & -2 \\
0 & -2 & 1 \\
-1 & 3 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & -2 \\
0 & -2 & 1 \\
-1 & 3 & 0
\end{array}\right]\)
∴ A^-1 = \(\left[\begin{array}{lll}
1 & 2 & -2 \\
0 & -2 & 1 \\
-1 & 3 & 0
\end{array}\right]\)
= 1\(\left|\begin{array}{ll}
-2 & 1 \\
3 & 0
\end{array}\right|\) – 2\(\left|\begin{array}{ll}
0 & 1 \\
-1 & 1
\end{array}\right|\) – 2\(\left|\begin{array}{ll}
0 & -2 \\
-1 & 3
\end{array}\right|\)
|A| = 1(0 – 3) – 2(0 + 1) – 2(0 – 2)
= -3 – 2 + 4
= -1 ≠ 0
∴ A^-1 exists.
We have
AA^-1 = I


∴ A^-1 = \(\left[\begin{array}{lll}
3 & 6 & 2 \\
1 & 2 & 1 \\
2 & 5 & 2
\end{array}\right]\)

Question 8.
Find the inverse of A = \(\left[\begin{array}{ccc}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{array}\right]\) by
(i) elementary row transformations
Solution:
|A| = \(\left|\begin{array}{ccc}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{array}\right|\)
= cosθ (cosθ – 0) + sinθ (sinθ – 0) + 0
= cos²θ + sin²θ = 1 ≠ 0
∴ A^-1 exists.
(i) Consider AA^-1 = I

(ii) elementary column transformations
Solution:
Consider A^-1A = I

Question 9.
If A = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 0 \\
3 & 1
\end{array}\right]\) find AB and (AB)^-1. Verify that (AB)^-1 = B^-1A^-1
Solution:
AB = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\) \(\left[\begin{array}{ll}
1 & 0 \\
3 & 1
\end{array}\right]\)




From (1) and (2), (AB)^-1 = B^-1 ∙ A^-1.

Question 10.
If A = \(\left[\begin{array}{ll}
4 & 5 \\
2 & 1
\end{array}\right]\), then show that A^-1 = \(\frac{1}{6}\)(A – 5I)
Solution:
|A| = \(\left|\begin{array}{ll}
4 & 5 \\
2 & 1
\end{array}\right|\) = 4 – 10 = -6 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

Question 11.
Find matrix X such that AX = B, where A = \(\left[\begin{array}{ll}
1 & 2 \\
-1 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
0 & 1 \\
2 & 4
\end{array}\right]\)
Solution:
AX = B

Question 12.
Find X, if AX = B where A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
-1 & 1 & 2 \\
1 & 2 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\).
Solution:
AX = B

Question 13.
If A = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ll}
4 & 1 \\
3 & 1
\end{array}\right]\) and C = \(\left[\begin{array}{ll}
24 & 7 \\
31 & 9
\end{array}\right]\) then find matrix X such that AXB = C.
Solution:
AXB = C
∴ \(\left(\begin{array}{ll}
1 & 1 \\
1 & 2
\end{array}\right)(\mathrm{XB})\) =\(\left[\begin{array}{ll}
24 & 7 \\
31 & 9
\end{array}\right]\)
First we perform the row transformations.

Question 14.
Find the inverse of \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\) by adjoint method.
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right|\)
= 1(7 – 20) – 2(7 – 10) + 3(4 – 2)
= -13 + 6 + 6 = -1 ≠ 0
∴ A^-1 exists.
First we have to find the cofactor matrix
= [A_ij]_3×3 where A_ij = (-1)^i+jM_ij
Now, A₁₁ = (-1)¹⁺¹M₁₁ = \(\left|\begin{array}{ll}
1 & 5 \\
4 & 7
\end{array}\right|\) = 7 – 20 = -13
A₁₂ = (-1)¹⁺²M₁₂ = \(\left|\begin{array}{ll}
1 & 5 \\
2 & 7
\end{array}\right|\) = -(7 – 10) = 3

Question 15.
Find the inverse of \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) by adjoint method.
Solution:
where A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\)
|A| = 1(2 – 6) – 0(0 – 3) + 1(0 – 2)
|A| = -4 – 2
|A| = -6 ≠ 0
∴ A^-1 exists.
First we have to find the cofactor matrix
= [A_ij]3×3, where A_ij = (-1)^i+jM_ij

Question 16.
Find A^-1 by adjoint method and by elementary transformations if A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
-1 & 1 & 2 \\
1 & 2 & 4
\end{array}\right]\)
Solution:
|A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
-1 & 1 & 2 \\
1 & 2 & 4
\end{array}\right|\)
= 1(4 – 4) – 2(-4 – 2) + 3(-2 – 1)
= 0 + 12 – 9 = 3 ≠ 0
∴ A^-1 exists.
A^-1by adjoint method :
We have to find the cofactor matrix
= [A_ij]_3×3, where A_ij = (-1)^i+j M_ij

Question 17.
Find the inverse of A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) by elementary column transformations.
Solution:
|A| = \(\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right|\)
= 1 (2 – 6) – 0 + 1 (0 – 2)
= -4 – 2= -6 ≠ 0
∴ A^-1 exists.
Consider A^-1A = I
∴ A^-1\(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
By C₃ – C₁, we get,

Question 18.
Find the inverse of \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\) by elementary row transformations.
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right|\)
= 1(7 – 20) – 2(7 – 10) + 3(4 – 2)
= -13 + 6 + 6 = -1 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

Question 19.
Show with usual notations that for any matrix A = [a_ij]_3×3
(i) a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃ = 0
Solution:
A = [a_ij]_3×3 = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\)
(i) A₂₁ = (-1)²⁺¹M₂₁ = \(-\left|\begin{array}{ll}
a_{12} & a_{13} \\
a_{32} & a_{33}
\end{array}\right|\)
= -(a₁₂a₃₃ – a₁₃a₃₂)
= -a₁₂a₃₃ + a₁₃a₃₂
A₂₂ = (-1)²⁺²M₂₂ = \(\left|\begin{array}{ll}
a_{11} & a_{13} \\
a_{31} & a_{33}
\end{array}\right|\)
= a₁₁a₃₃ – a₁₃a₃₁
A₂₃ = (-1)²⁺³M₂₃ = \(-\left|\begin{array}{ll}
a_{11} & a_{12} \\
a_{31} & a_{32}
\end{array}\right|\)
= -(a₁₁a₃₂ – a₁₂a₃₁)
= -a₁₁a₃₂+ a₁₂a₃₁
∴ a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃
= a₁₁(-a₁₂₃₃ + a₁₃a₃₂) + a₁₂(a₁₁a₃₃ – a₁₃a₃₁) + a₁₃(-a₁₁a₃₂ + a₁₂a₃₁)
= -a₁₁a₁₂a₃₃ + a₁₁a₁₃a₃₂ + a₁₁a₁₂a₃₃ – a₁₂a₁₃a₃₁ – a₁₁a₁₃a₃₂ + a₁₂a₁₃a₃₁
= 0

(ii) a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃ = |A|
Solution:

Question 20.
If A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\), then find a matrix X such that XA= B.
Solution:
Consider XA = B