Class 11

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

iv.

= 1
= R.H.S

v.

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

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

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

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

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

iii. tan 105° = tan (60° +45°)

iv. cot 225°

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

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

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

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

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

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

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

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

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

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

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

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

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

= tan (45° + 27°)
= tan 72° = R.H.S

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

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

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

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

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

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

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

iii.

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

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

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

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

I. Select the correct option from the given alternatives.

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

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

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

Explanation:

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

Explanation:

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

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

Explanation:

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

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

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

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

Explanation:

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

Explanation:

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

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

Explanation:

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

Explanation:

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

II. Answer the following:

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

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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 10.
Prove the following:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Alternate Method:

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

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

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

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

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

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

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

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

∴ 16x² – 9y² = 144

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

16x² – 9y² = 576

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

ii. We know that,

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

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

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

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

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.4

Question 1.
Find the slope, x-intercept, y-intercept of each of the following lines, i. 2x + 3y-6 = 0 ii. 3x-y-9 = 0 iii. x + 2y = 0
Solution:
i. Given equation of the line is 2x + 3y – 6 = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 2, b = 3, c = -6

ii. Given equation of the line is 3x – y – 9 = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 3, b = – 1, c = – 9

iii. Given equation of the line is x + 2y = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 1, b = 2, c = 0

Question 2.
Write each of the following equations in ax + by + c = 0 form.
i. y = 2x – 4
ii. y = 4
iii. \(\frac{x}{2}+\frac{y}{4}=1\)
iv. \(\frac{x}{3}-\frac{y}{2}=0\)
i. y = 2x – 4
∴ 2x – y – 4 = 0 is the equation in ax + by + c = 0 form.

ii. y = 4
∴ 0x + 1y – 4 = 0 is the equation in ax + by + c = 0 form.

iii. \(\frac{x}{2}+\frac{y}{4}=1\)
∴ \(\frac{2 x+y}{4}\)
∴ 2x + y – 4 = 0 is the equation in ax + by + c = 0 form.

iv. \(\frac{x}{3}-\frac{y}{2}=0\)
∴ 2x – 3y = 0
∴ 2x – 3y + 0 = 0 is the equation in ax + by + c = 0 form.
[Note: Answer given in the textbook is ‘2x – 3y – 6 = 0’. However, as per our calculation it is ‘2x-3y + 0 = 0’.]

Question 3.
Show that the lines x – 2y – 7 = 0 and 2x – 4y + 15 = 0 are parallel to each other.
Solution:
Let m₁ be the slope of the line x – 2y – 7 = 0.
∴ m₁ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-1}{-2}=\frac{1}{2}\)
Let m₂ be the slope of the line 2x – 4y + 15 = 0.
∴ m₂ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-2}{-4}=\frac{1}{2}\)
Since m₁ = m₂
the given lines are parallel to each other.

Question 4.
Show that the lines x – 2y – 7 = 0 and 2x + y + 1 = 0 are perpendicular to each other. Find their point of intersection.
Solution:
Let m₁ be the slope of the line x – 2y – 7 = 0.
∴ m₁ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-1}{-2}=\frac{1}{2}\)
Let m₂ be the slope of the line 2x + y + 1 = 0.
∴ m₂ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-2}{1}=-2\)
Since m₁ x m₁ = \(\frac{1}{2}\) x (- 2) = -1,
the given lines are perpendicular to each other. Consider,
x – 2y – 7 = 0 …(i)
2x + y + 1 =0 …(ii)
Multiplying equation (ii) by 2, we get
4x + 2y + 2 = 0 …(iii)
Adding equations (i) and (iii), we get
5x – 5 = 0
∴ x = 1
Substituting x = 1 in equation (ii), we get
2 + y + 1 = 0
∴ y = – 3
∴ The point of intersection of the given lines is (1,-3).

Question 5.
If the line 3x + 4y = p makes a triangle of area 24 square units with the co-ordinate axes, then find the value of p.
Solution:
Let the line 3x + 4y = p cuts the X and Y axes at points A and B respectively.
3x + 4y = p

This equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\),
where a = \(\frac{p}{3}\) and b = \(\frac{p}{4}\)
∴ A (a, 0) ≡ (\(\frac{p}{3}\), 0) and B ≡ (0, b) = (0, \(\frac{p}{4}\))
∴ OA = \(\frac{p}{3}\) and OB = \(\frac{p}{4}\)
Given, A (∆OAB) = 24 sq. units
∴ \(\left|\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\right|=24\)
∴ \(\left|\frac{1}{2} \times \frac{\mathrm{p}}{3} \times \frac{\mathrm{p}}{4}\right|=24\)
∴ p² = 576
∴ p = ± 24

Question 6.
Find the co-ordinates of the foot of the perpendicular drawn from the point A(- 2,3) to the line 3x-y -1 = 0.
Solution:
Let M be the foot of perpendicular drawn from
point A(- 2,3) to the line
3x-y- 1 = 0 …(i)
Slope of the line 3x-y – 1 = 0 is \(\frac{-3}{-1}\) =3.
Since AM ⊥ to line (i),
slope of AM = \(\frac{-1}{3}\)
∴ Equation of AM is
y – 3 = \(\frac{-1}{3}\)(x + 2)

∴ 3(y – 3) = – 1(x + 2)
∴ 3y – 9 = -x – 2
∴ x + 3y – 7 = 0 …………(ii)
The foot of perpendicular i.e., point M, is the point of intersection of equations (i) and (ii).
By (i) x 3 + (ii), we get 10x -10 = 0
∴ x = 1
Substituting x = 1 in (ii), we get
1 + 3y – 7 = 0
∴ 3y = 6
∴ y = 2
∴ The co-ordinates of the foot of the perpendicular Mare (1,2).

Question 7.
Find the co-ordinates of the circumcentre of the triangle whose vertices are A(- 2, 3), B(6, -1), C(4,3),e
Solution:
Here, A(-2, 3), B(6, -1), C(4, 3) are the vertices of ∆ABC.
Let F be the circumcentre of AABC.
Let FD and FE be the perpendicular bisectors of the sides BC and AC respectively.
∴ D and E are the midpoints of side BC and AC respectively.

Since FD passes through (5, 1) and has slope 1/2 equation of FD is
y – 1 = \(\frac{1}{2}\)(x-5)
∴ 2 (y – 1) = x – 5
∴ 2y – 2 = x – 5
∴ x – 2y – 3 = 0 …(i)
Since both the points A and C have same y co-ordinates i.e. 3,
the given points lie on the line y = 3.
Since the equation FE passes through E(1, 3),
the equation of FE is x = 1. .. .(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value of x in (i), we get
1 – 2y -3 = 0
∴ y = -1
∴ Co-ordinates of circumcentre F ≡ (1, – 1).

Question 8.
Find the co-ordinates of the orthocentre of the triangle whose vertices are A(3, – 2), B(7,6), C (-1,2).
Solution:
Let O be the orthocentre of ∆ABC.
Let AD and BE be the altitudes on the sides BC and AC respectively.

Slope of side BC = \(\frac{2-6}{-1-7}=\frac{-4}{-8}=\frac{1}{2}\)
∴ Slope of AD = – 2 [∵ AD ⊥ BC]
∴ Equation of line AD is
y – (-2) = (- 2) (x – 3)
∴ y + 2 = -2x + 6
∴ 2x + y -4 = 0 …(i)
Slope of side AC = \(\frac{-2-2}{3-(-1)}=\frac{-4}{4}\) = -1
∴ Slope of BE = 1 …[ ∵ BE ⊥ AC]
∴ Equation of line BE is
y – 6 = 1(x – 7)
∴ y – 6 = x – 1
∴ x = y + 1 …(ii)
Substituting x = y + 1 in (i), we get
2(y + 1) + y – 4 = 0
∴ 2y + 2 + y – 4 = 0
∴ 3y – 2 = 0
∴ y = \(\frac{2}{3} in (ii), we get
Substituting y = [latex]\frac{2}{3}\) in (ii), we get
x = \(\frac{2}{3}+1=\frac{5}{3}\)
∴ Co-ordinates of orthocentre, O = \(\left(\frac{5}{3}, \frac{2}{3}\right)\)

Question 9.
Show that the lines 3 – 4y + 5 = 0, lx – 8y + 5 = 0 and 4JC + 5y – 45 = 0 are concurrent. Find their point of concurrence.
Solution:
The number of lines intersecting at a point are called concurrent lines and their point of intersection is called the point of concurrence. Equations of the given lines are
3x – 4y + 5 = 0 …(i)
7x-8y + 5 = 0 …(ii)
4x + 5y – 45 = 0 …(iii)
By (i) x 2 – (ii), we get
– x + 5 = 0
∴ x = 5
Substituting x = 5 in (i), we get
3(5) – 4y + 5 = 0
∴ -4y = – 20
∴ y = 5
∴ The point of intersection of lines (i) and (ii) is given by (5, 5).
Substituting x = 5 and y = 5 in L.H.S. of (iii), we get
L.H.S. = 4(5) + 5(5) – 45
= 20 + 25 – 45
= 0
= R.H.S.
∴ Line (iii) also passes through (5, 5).
Hence, the given three lines are concurrent and the point of concurrence is (5, 5).

Question 10.
Find the equation of the line whose x-intercept is 3 and which ¡s perpendicular to the line 3x – y + 23 = 0.
Solution:
Slope of the line 3x – y + 23 = 0 is 3.
∴ Slope of the required line perpendicular to
3x – y + 23 = 0 is \(\frac{-1}{3}\)
Since the x-intercept of the required line is 3, it passes through (3, 0).
∴ The equation of the required line is ‘
y – 0 = \(\frac{-1}{3}\)(x – 3)
∴ 3y = x + 3
∴ x + 3y = 3

Question 11.
Find the distance of the origin from the line 7x + 24y – 50 = 0.
Solution:
Let p be the perpendicular distance of origin
fromtheline7x + 24y – 50 = 0
Here, a = 7, b = 24, c = -50

Question 12.
Find the distance of the point A(- 2, 3) from the line 12x – 5y – 13 = 0.
Solution:
Let p be the perpendicular distance of the point A(- 2, 3) from the line 12x – 5y – 13 = 0
Here, a = 12, b = – 5, c = – 13, x₁ = -2, y₁ = 3

Question 13.
Find the distance between parallel lines 4x – 3y + 5 = 0 and 4xr – 3y + 7 = 0.
Solution:
Equations of the given parallel lines are 4x – 3y + 5 = 0 and 4x – 3y + 1 = 0
Here, a = 4, b = – 3, c₁ = 5 and c₂ = 7
∴ Distance between the parallel lines

Question 14.
Find the distance between the parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.
Solution:
Equations of the given parallel lines are 3x + 2y + 6 = 0 and
9x + 6y – 1 = 0 i.e., 3x + 2y – \(\frac{7}{3}\) =0
Here, a = 3, b = 2, c₁ = 6 and c₂ = \(\frac{-7}{3}\)
∴ Distance between the parallel lines

Question 15.
Find the points on the line x + y – 4 = 0 which are at a unit distance from the line 4JC + 3y = 10.
Solution:
Let P(x₁, y₁) be a point on the line x + y – 4 = o.
∴ x₁ + y₁ – 4 = 0
∴ y₁ = 4 – x₁ …(i)
Also, distance of P from the line 4x + 3y- 10 = 0 is 1

∴ 5 = | x₁ + 2 |
∴ x₁ + 2 = ± 5
∴ x₁ + 2 = 5 or x₁ + 2 = – 5
∴ x₁ = 3 or x₁ = – 7
From (i), when x₁ = 3, y₁ = 1
and when x₁ = -7, y₁ = 11
∴ The required points are (3, 1) and (-7, 11).
[Note: The question has been modified]

Question 16.
Find the equation of the line parallel to the X-axis and passing through the point of intersection of lines x + y – 2 = 0 and 4x + 3y = 10.
Solution:
Let u = x + y – 2 = 0 and v = 4x + 3y – 10 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
∴ (x + y – 2) + k(4x + 3y – 10) = 0 …(i)
∴ x + y – 2 + 4kx + 3ky – 10k = 0
∴ x + 4kx + y + 3ky – 2 – 10k = 0
∴ (1+ 4k)x + (1 + 3k)y – 2 – 10k = 0
But, this line is parallel to X-axis.
∴ Its slope = 0
∴ \(\frac{-(1+4 k)}{1+3 k}\) = 0
∴ 1 + 4k = 0
∴ k = \(\frac{-1}{4}\)
Substituting the value of k in (i), we get
(x + y – 2) + (4x + 3y – 10) = 0
∴ 4(x +y – 2) – (4x + 3y -10 ) = 0
∴ 4x + 4y – 8 – 4x – 3y + 10 = 0
∴ y + 2 = 0, which is the equation of the required line.
[Note: Answer given in the textbook is 5y – 8= 0. However, as per our calculation it is y + 2 = 0.]

Question 17.
Find the equation of the line passing through the point of intersection of lines x + y – 2 = 0 and 2xr – 3y + 4 = 0 and making intercept 3 on the X-axis.
Solution:
Let u ≡ x + y – 2 = 0 and v ≡ 2x – 3y + 4 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
∴ (x +y – 2) + k(2x – 3y + 4) = 0 …(i)
But, x-intercept of line is 3.
∴ It passes through (3, 0).
Substituting x = 3 and y = 0 in (i), we get
(3 + 0 – 2) + k(6 – 0 + 4) = 0
∴ 1 + 10k = 0
k = \(\frac{-1}{10}\)
Substituting the value of k in (i), we get (x + y – 2) + \(\left(\frac{-1}{10}\right)\) (2x – 3y + 4) = 0
∴ 10(x + y – 2) – (2x – 3y + 4) = 0
∴ 10x + 10y -20 — 2x + 3y-4 = 0
∴ 8x + 13y – 24 = 0, which is the equation of the required line.

Question 18.
If A(4, 3), B(0, 0) and C(2, 3) are the vertices of ΔABC, then find the equation of bisector of angle BAC.
Solution:
Let the bisector of ∠ BAC meets BC at point D.
∴ Point D divides seg BC in the ratio l(AB) : l(AC)

∴ 18 (y – 3) = 6 (x – 4)
∴ 3(y – 3) = x – 4
∴ 3y – 9 = x – 4
∴ x – 3y + 5 = 0

Question 19.
D(- 1, 8), E(4, – 2), F(- 5, – 3) are midpoints of sides BC, CA and AB of AABC. Find
i. equations of sides of ΔABC.
ii. co-ordinates of the circumcentre of ΔABC.
Solution:
Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of ΔABC.
Given, points D, E and F are midpoints of sides BC, CA and AB respectively of ΔABC.

∴ x₁ + x₂ = -10 …………. (v)
and y₁ + y₂ = – 6 …………(vi)
For x-coordinates:
Adding (i), (iii) and (v), we get
2x₁ + 2x₂ + 2x₃ = – 4
∴ x₁ + x₂ + x₃ = -2 …………..(vii)
Solving (i) and (vii), we get x₁ = 0
Solving (iii) and (vii), we get x₂ = – 10
Solving (v) and (vii), we get x₃ = 8

For y-coordinates:
Adding (ii), (iv) and (vi), we get 2y₁ + 2y₂ + 2y₃ = 6
y₁ + y₂ + y₃ = 3 …….(viii)
Solving (ii) and (viii), we get y₁ = -13
Solving (iv) and (viii), we get y₂ = 7
Solving (vi) and (viii), we get y₃ = 9
∴ Vertices of AABC are A(0, – 13), B(- 10, 7), C(8, 9)

∴ 8(y + 13) = 22x
∴ 4(y + 13) = 11x
∴ 11x – 4y – 52 = 0

ii. Here, A(0, – 13), B(- 10, 7), C(8, 9) are the vertices of ΔABC.
Let F be the circumcentre of AABC.
Let FD and FE be perpendicular bisectors of the sides BC and AC respectively.
D and E are the midpoints of side BC and AC.

∴ Slope of FD = -9 … [ ∵ FD ⊥ BC]
Since FD passes through (-1, 8) and has slope -9, equation of FD is
y – 8 = -9 (x +1)
∴ y – 8 = -9x – 9
∴ y = -9x – 1
Also, slope of AC = \(\frac{-13-9}{0-8}=\frac{11}{4}\)
∴ Slope of FE = \(\frac{-4}{11}\) [ ∵ FE ⊥ AC]
Since FE passes through (4, -2) and has slope -4
\(\frac{-4}{11}\), equation of FE is
(y + 2) = \(\frac{-4}{11}\) (x – 4)
∴ 11(y + 2) = -4(x – 4)
∴ 11y + 22 = – 4x + 16
∴ 4x + 11y = -6 …………(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value ofy in (ii), we get
4x + 11(-9x- 1) = – 6
∴ 4x – 99x -11 = – 6
∴ -95x = 5
∴ x = \(\frac{-1}{19}\)
Substituting the value of x in (i), we get
y = -9(\(\frac{-1}{19}\)) – 1 = \(\frac{-10}{19}\)
∴ Co-ordinates of circumcentre F ≡ \(\left(\frac{-1}{19}, \frac{-10}{19}\right)\)

Question 20.
0(0, 0), A(6, 0) and B(0, 8) are vertices of a triangle. Find the co-ordinates of the incentre of ∆OAB.
Solution:
Let bisector of ∠O meet AB at point D and bisector of ∠A meet BO at point E
∴ Point D divides seg AB in the ratio l(OA): l(OB)
and point E divides seg BO in the ratio l(AB): l(AO)
Let I be the incentre of ∠OAB.

∴ Point D divides AB internally in 6 : 8
i.e. 3 :4

∴ y = x …(i)
Now, by distance formula,
l(AB) = \(\begin{aligned}
&=\sqrt{(6-0)^{2}+(0-8)^{2}} \\
&=\sqrt{36+64}=10
\end{aligned}\)
l(AO) = \(\sqrt{(6-0)^{2}+(0-0)^{2}}\) = 6
∴ Point E divides BO internally in 10 : 6 i.e. 5:3

∴ -2y = x – 6
∴ x + 2y = 6 …(ii)
To find co-ordinates of incentre, we have to solve equations (i) and (ii).
Substituting y = x in (ii), we get
x + 2x = 6
∴ x = 2
Substituting the value of x in (i), we get
y = 2
∴ Co-ordinates of incentre I ≡ (2, 2).

Alternate Method:
Let I be the incentre.
I lies in the 1st quadrant.
OPIR is a square having side length r.
Since OA = 6, OP = r,
PA = 6 – r
Since PA = AQ,
AQ = 6 – r …(i)
Since OB = 8, OR = r,
BR = 8 – r
∴ BR = BQ
∴ BQ = 8 – r …(ii)

AB = BQ + AQ
Also, AB = \(\begin{aligned}
&=\sqrt{\mathrm{OA}^{2}+\mathrm{OB}^{2}} \\
&=\sqrt{6^{2}+8^{2}} \\
&=\sqrt{100}=10
\end{aligned}\)
∴ BQ + AQ= 10
∴ (8 – r) + (6 – r) = 10
∴ 2r = 14- 10 = 4
∴ r = 2
∴ I = (2,2)

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.3

Question 1.
Write the equation of the line:
i. parallel to the X-axis and at a distance of 5 units from it and above it.
ii. parallel to the Y-axis and at a distance of 5 units from it and to the left of it.
iii. parallel to the X-axis and at a distance of 4 units from the point (- 2,3).
Solution:
i. Equation of a line parallel to X-axis is y = k. Since the line is at a distance of 5 units above the X-axis, k = 5
∴ The equation of the required line is y = 5.

ii. Equation of a line parallel to the Y-axis is x = h. Since the line is at a distance of 5 units to the left of the Y-axis, h = -5
∴ The equation of the required line is x = -5.
[Note: Answer given in the textbook is ‘y = -5
However, we found that ‘x = – 5’.]

iii. Equation of a line parallel to the X-axis is of the form y = k (k > 0 or k < 0).
Since the line is at a distance of 4 units from the point (- 2, 3),
k = 4 + 3 = 7 or k = 3- 4 = -1
∴ The equation of the required line is y = 1 or y = – 1.

Question 2.
Obtain the equation of the line:
i. parallel to the X-axis and making an intercept of 3 units on the Y-axis.
ii. parallel to the Y-axis and making an intercept of 4 units on the X-axis.
Solution:
i. Equation of a line parallel to X-axis with y-intercept ‘k’ isy = k.
Here, y-intercept = 3
∴ The equation of the required line is y = 3.

ii. Equation of a line parallel to Y-axis with x-intercept ‘h’ is x = h.
Here, x-intercept = 4
∴ The equation of the required line is x = 4.

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

ii. Equation of a line parallel to X-axis is of the formy = k.
Since the line passes through B(4, – 3), k = -3
∴ The equation of the required line is y = – 3.

Question 4.
Find the equation of the line:
i. passing through the points A(2, 0) and B(3,4)
ii. passing through the points P(2, 1) and Q(2,-1)
Solution:
i. The required line passes through the points A(2, 0) and B(3,4).
Equation of the line in two point form is \(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
Here, (x₁y₁) = (2,0) and (x₁,y₂) = (3,4)
∴ The equation of the required line is
∴ \(\frac{y-0}{4-0}=\frac{x-2}{3-2}\)
∴ \(\frac{y}{4}=\frac{x-2}{1}\)
∴ y = 4(x – 2)
∴ y = 4x – 8
∴ 4x – y – 8 = 0

ii. The required line passes through the points P(2, 1) and Q(2,-1).
Since both the given points have same
x co-ordinates i.e. 2,
the given points lie on the line x = 2.
∴ The equation of the required line is x = 2.

Question 5.
Find the equation of the line:
i. containing the origin and having inclination 60°.
ii. passing through the origin and parallel to AB, where A is (2,4) and B is (1,7).
iii. having slope 1/2 and containing the point (3, -2)
iv. containing the point A(3, 5) and having slope 2/3
v. containing the point A(4, 3) and having inclination 120°.
vi. passing through the origin and which bisects the portion of the line 3JC + y = 6 intercepted between the co-ordinate axes.
Solution:
i. Given, Inclination of line = θ = 60°
Slope of the line (m) = tan θ = tan 60°
= \(\sqrt{3}\)
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
.‘. The equation of the required line is y = \(\sqrt{3}\) x

ii. Given, A (2, 4) and B (1, 7)
Slope of AB = \(\frac{7-4}{1-2}\) = -3 1-2
Since the required line is parallel to line AB, slope of required line (m) = slope of AB
∴ m = – 3 and the required line passes through the origin.
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is y = – 3x

iii. Given, slope(m) = \(=\frac{1}{2}\) and the line passes through (3, – 2).
Equation of the line in slope point form is
y-y ₁= m(x-x₁)
∴ The equation of the required line is
[y-(- 2)]=\(\frac{1}{2}\)(x-3)
∴ 2(y + 2)=x – 3
∴ 2y + 4 = x – 3
∴ x – 2y – 7 = 0

iv. Given, slope(m) = \(\frac{2}{3}\) and the line passes through (3, 5).
Equation of the line in slope point form is y-y₁ = m(x -x₁)
∴ The equation of the required line is y – 5 = \(\frac{2}{3}\)(x-3)
∴ 3 (y – 5) = 2 (x – 3)
∴ 3y – 15 = 2x – 6
∴ 2x – 3y + 9 = 0

v. Given, Inclination of line = θ = 120°
Slope of the line (m) = tan θ = tan 120°
= tan (90° + 30°)
= – cot 30°
= – \(\sqrt{3}\)
and the line passes through A(4, 3).
Equation of the line in slope point form is y-y₁ = m(x -x₁)
∴ The equation of the required line is
y- 3 = –\(\sqrt{3}\)(x-4)
∴ y – 3 = –\(\sqrt{3}\) x + 4\(\sqrt{3}\)
∴ \(\sqrt{3}\)x + y – 3 -4\(\sqrt{3}\) = 0

vi.

Given equation of the line is 3x +y = 6.
∴ \(\frac{x}{2}+\frac{y}{6}=1\)
This equation is of the form \(\frac{x}{\mathrm{a}}+\frac{y}{\mathrm{~b}}\) = 1,
where a = 2, b = 6
∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2, 0) and B(0, 6) respectively. Required line is passing through the midpoint of AB.
∴ Midpoint of AB = ( \(\frac{2+0}{2}, \frac{0+6}{2}\) ) = (1,3)
∴ Required line passes through (0, 0) and (1,3).
Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the required line is
\(\frac{y-0}{3-0}=\frac{x-0}{1-0}\)
\(\frac{y}{3}=\frac{x}{1}\)
∴ y = 3x
∴ 3x – y = 0

Alternate Method:
Given equation of the line is 3x + y = 6 …(i)
Substitute y = 0 in (i) to get a point on X-axis.
∴ 3x + 0 = 6
∴ x = 2
Substitute x = 0 in (i) to get a point on Y-axis.
∴ 3(0) + 7 = 6
∴ y = 6
∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2,0) and B(0,6) respectively.
Let M be the midpoint of AB.
M = \(\left(\frac{2+0}{2}, \frac{0+6}{2}\right)\) = (1,3)
Slope of OM (m) = \(\frac{3-0}{1-0}\) = 3
Equation of OM is of the formy = mx.
∴ The equation of the required line is y = 3x
∴ 3x – y = 0

Question 6.
Line y = mx + c passes through the points A(2,1) and B(3,2). Determine m and c.
Solution:
Given, A(2, 1) and B(3,2)
Equation of the line in two point form is \(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the required line is
\(\frac{y-1}{2-1}=\frac{x-2}{3-2}\)
∴ \(\frac{y-1}{1}=\frac{x-2}{1}\)
∴ y – 1 = x – 2
∴ y = x – 1
Comparing this equation with y = mx + c, we get
m = 1 and c = – 1

Alternate Method:
Points A(2, 1) and B(3, 2) lie on the line y = mx + c.
∴ They must satisfy the equation.
∴ 2m + c = 1 …(i)
and 3m + c = 2 …(ii)
equation (ii) – equation (i) gives m = 1
Substituting m = 1 in (i), we get 2(1) + c = 1
∴ c = 1 – 2 = – 1

Question 7.
Find the equation of the line having inclination 135° and making x-intercept 7.
Solution:
Given, Inclination of line = 0 = 135°
∴ Slope of the line (m) = tan 0 = tan 135°
= tan (90° + 45°)
= – cot 45° = – 1 x-intercept of the required line is 7.
∴ The line passes through (7, 0).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
∴ The equation of the required line is y — 0 = – 1 (x – 7)
∴ y = -x + 7
∴ x + y – 7 = 0

Question 8.
The vertices of a triangle are A(3, 4), B(2, 0) and C(- 1, 6). Find the equations of the lines containing
i. side BC
ii. the median AD
iii. the midpoints of sides AB and BC.
Solution:
Vertices of AABC are A(3, 4), B(2, 0) and C(- 1, 6).
i. Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the side BC is
\(\frac{y-0}{6-0}=\frac{x-2}{-1-2}\)
\(\frac{y}{6}=\frac{x-2}{-3}\)
∴ – 3y = 6x – 12
∴ 6x + 3y – 12 = 0
∴ 2x + y – 4 = 0

ii. Let D be the midpoint of side BC.
Then, AD is the median through A.
∴ D = \(\left(\frac{2-1}{2}, \frac{0+6}{2}\right)=\left(\frac{1}{2}, 3\right)\)
The median AD passes through the points
A(3,4) and D( \(\frac{1}{2}\) , 3)

∴ The equation of the median AD is
\(\frac{y-4}{3-4}=\frac{x-3}{\frac{1}{2}-3}\)
\(\frac{y-4}{-1}=\frac{x-3}{-\frac{5}{2}}\)
\(\frac{5}{2}\)(y-4) = x – 3
∴ 5y – 20 = 2x – 6
∴ 2x – 5y + 14 = 0

iii. Let D and E be the midpoints of side AB and side BC respectively.
The equation of the line DE is

∴ -4(y-2) = 2x-5
∴ 2x + 4y – 13 = 0

Question 9.
Find the x and y-intercepts of the following lines:
i. \(\frac{x}{3}+\frac{y}{2}=1\)
ii. \(\frac{3 x}{2}+\frac{2 y}{3}=1\)
iii. 2x – 3y + 12 = 0
Solution:
i. Given equation of the line is latex]\frac{x}{3}+\frac{y}{2}=1[/latex]
This is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = 3, y-intercept = 2

ii. Given equation of the line is \(\frac{3 x}{2}+\frac{2 y}{3}\) = 1
∴ \(\frac{x}{\left(\frac{2}{3}\right)}+\frac{y}{\left(\frac{3}{2}\right)}\) = 1
This is of the form = \(\frac{x}{a}+\frac{y}{b}\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = \(\frac{2}{3}\) and y-intercept = \(\frac{3}{2}\)

iii. Given equation of the line is 2x – 3y + 12 = 0
∴ 2x – 3y = – 12
∴ \(\frac{2 x}{(-12)}-\frac{3 y}{(-12)}=1\)
∴ \(\frac{x}{-6}+\frac{y}{4}=1\)
This is of the form \(\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = – 6 and y-intercept = 4

Question 10.
Find equations of the line which contains the point A(l, 3) and the sum of whose intercepts on the co-ordinate axes is zero.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be
\(\frac{x}{a}+\frac{y}{b}=1\) ………..(i)
Since, the sum of the intercepts of the line is zero.
∴ a + b = 0
∴ b = – a
Substituting b = – a in (i), we get
\(\frac{x}{a}+\frac{y}{(-a)}=1\)
x – y = a .. .(ii)
Since, the line passes through A(1, 3).
∴ 1 – 3 = a
∴ a = – 2
Substituting the value of a in (ii), equation of the required line is
∴ x – y = – 2,
∴ x – y + 2 = 0

Case II: Line passing through origin.
Slope of line passing through origin and
A(1, 3) is m = \(\frac{3-0}{1-0}\) = 3
∴ Equation of the line having slope m and passing through origin (0, 0) is / = mx.
∴ The equation of the required line is y = 3x
∴ 3x – y = 0

Question 11.
Find equations of the line containing the point A(3, 4) and making equal intercepts on the co-ordinate axes.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}=1\) …………(i)
This line passes through A(3, 4).
∴ \(\frac{3}{a}+\frac{4}{b}=1\)……………..(ii)
Since, the required line make equal intercepts on the co-ordinate axes.
∴ a = b …(iii)
Substituting the value of b in (ii), we get
\(\frac{3}{a}+\frac{4}{a}=1\)
∴ \(\frac{7}{a}=1\)
∴ a = 7
∴ b = 7 …[From (iii)]
Substituting the values of a and b in (i), equation of the required line is
\(\frac{x}{7}+\frac{y}{7}=1\) = 1
∴ x + y = 7

Case II: Line passing through origin.
Slope of line passing through origin and A(3,4) is m = \(=\frac{4-0}{3-0}=\frac{4}{3}\)
∴ Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is 4
y = \(\frac{4}{3}\)x
∴ 4x – 3y = 0

Question 12.
Find the equations of the altitudes of the triangle whose vertices are A(2, 5), B(6, – 1 ) and C(- 4, – 3).
Solution:

A(2, 5), B(6, – 1), C(- 4, – 3) are the vertices of ∆ABC.
Let AD, BE and CF be the altitudes through the vertices A, B and C respectively of ∆ABC.
∴ Slope of AD = -5 …[∵AD ⊥ BC]
Since altitude AD passes through (2, 5) and has slope – 5,
equation of the altitude AD is y – 5 = -5 (x – 2)
∴ y – 5 = – 5x + 10
∴ 5x +y -15 = 0
Now, slope of AC = \(\frac{-3-5}{-4-2}=\frac{-8}{-6}=\frac{4}{3}\)
Slope of BE = \(\frac{-3}{4}\)
…[∵ BE ⊥ AC]
Since altitude BE passes through (6,-1) and has slope \(\frac{-3}{4}\),
equation of the altitude BE is
y-(-1) = \(\frac{-3}{4}\) (x – 6)
∴ 4 (y + 1) = – 3 (x – 6)
∴ 4y + 4 =-3x+ 18
∴ 3x + 4y – 14 = 0
Also, slope of AB = \(\frac{-1-5}{6-2}=\frac{-6}{4}=\frac{-3}{2}\)
∴ Slope of CF = \({2}{3}\) ….[∵ CF ⊥ AB]
Since altitude CF passes through (- 4, – 3) and has slope , \(\frac{2}{3}\)
equation of the altitude CF is
y-(-3) = \(\frac{2}{3}\)[x-(-4)]
∴ 3 (y + 3) = 2 (x + 4)
∴ 3y + 9 = 2x + 8
∴ 2x – 3y – 1 = 0

Question 13.
Find the equations of perpendicular bisectors of sides of the triangle whose vertices are P(-1, 8), Q(4, – 2) and R(- 5, – 3).
Solution:

Let A, B and C be the midpoints of sides PQ, QR and PR respectively of APQR.
A is the midpoint of side PQ.

Slope of perpendicular bisector of PQ is \(\frac{1}{2}\) and it passes through (\(\frac{3}{2}\)), 3).
Equation of the perpendicular bisector of side PQ is
y – 3 = \(\frac{1}{2}\)(x – \(\frac{3}{2}\))
y – 3 = (\(\frac{1}{2}\left(\frac{2 x-3}{2}\right)\))
∴ 4(y – 3) = 2x – 3
∴ 4y – 12 = 2x – 3
∴ 2x – 4y + 9 = 0
B is the midpoint of side QR
∴ B = \(\left(\frac{4-5}{2}, \frac{-2-3}{2}\right)=\left(\frac{-1}{2}, \frac{-5}{2}\right)\)
Slope of side QR = \(\frac{-3-(-2)}{-5-4}=\frac{-1}{-9}=\frac{1}{9}\)
∴ Slope of perpendicular bisector of QR is -9 and it passes through \(\left(-\frac{1}{2},-\frac{5}{2}\right)\)
∴ Equation of the perpendicular bisector of side QR is

∴ 2y + 5 = -18x – 9
∴ 18x + 2y + 14 = 0
∴ 9x + y + 7 = 0
C is the midpoint of side PR.

Equation of the perpendicular bisector of PR is \(y-\frac{5}{2}=-\frac{4}{11}(x+3)\)
∴ \(11\left(\frac{2 y-5}{2}\right)\) =-4(x + 3)
∴ 11(2y – 5) = – 8 (x + 3)
∴ 22y – 55 = – 8x – 24
∴ 8x + 22y -31 = 0

Question 14.
Find the co-ordinates of the orthocentre of the triangle whose vertices are A(2, – 2), B(l, 1) and C(-1,0).
Solution:
Let O be the orthocentre of AABC.
Let AM and BN be the altitudes of sides BC and AC respectively.
Now, slope of BC = \(\frac{0-1}{-1-1}=\frac{-1}{-2}=\frac{1}{2}\)
Slope of AM = -2 ,..[∵ AM ⊥ BC]
Since AM passes through (2, – 2) and has slope -2,
equation of the altitude AM is y – (- 2) = – 2 (x – 2)
∴ y + 2 = -2x + 4
∴ 2x + y – 2 = 0 …(i)

Also, slope of AC = \(\frac{0-(-2)}{-1-2}=\frac{2}{-3}\)
∴ Slope of BN = \(\frac{3}{2}\) …[∵ BN ⊥ AC]
Since BN passes through (1,1) and has slope \(\frac{3}{2}\), equation of the altitude BN is
y – 1 = \(\frac{3}{2}\)(x-1)
∴ 2y – 2 = 3x – 3
∴ 3x – 2y – 1 = 0 …(ii)
To find co-ordinates of orthocentre, we have to solve equations (i) and (ii).
By (i) x 2 + (ii), we get
7x – 5 = 0
∴ x = \(\frac{5}{7}\)
substituting x = \(\frac{5}{7}\) in eq (i), we get
2(\(\frac{5}{7}\)) + y – 2 = 0
∴ y = -2(\(\frac{5}{7}\)) + 2
∴ y = \(\frac{-10+14}{7}=\frac{4}{7}\)
∴ Coordinates of orthocentre O = \(\left(\frac{5}{7}, \frac{4}{7}\right)\)

Question 15.
N(3, – 4) is the foot of the perpendicular drawn from the origin to line L. Find the equation of line L.
Solution:

Slope of ON = \(\frac{-4-0}{3-0}=\frac{-4}{3}\)
Since line L ⊥ ON,
slope of the line L is \(\frac{3}{4}\) and it passes through point N(3, -4).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
Equation of line L is
y-(-4) = \(\frac{3}{4}\)(x-3)
∴ 4(y + 4) = 3(x – 3)
∴ 4y + 16 = 3x – 9
∴ 3x – 4y – 25 = 0

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.2

Question 1.
Find the slope of each of the lines which passes through the following points:
i. A(2, -1), B(4,3)
ii. C(- 2,3), D(5, 7)
iii. E(2,3), F(2, – 1)
iv. G(7,1), H(- 3,1)
Solution:
i. Here, A = (2, -1) andB = (4, 3)
Slope of line AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-(-1)}{4-2}=\frac{4}{2}\) = 2

ii. Here, C = (-2, 3) and D = (5, 7)
Slope of line CD = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{7-3}{5-(-2)}=\frac{4}{7}\)

iii. Here, E s (2, 3) and F = (2, -1)
Slope of line EF = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-1-3}{2-2}=\frac{-4}{0}\), which ix not defined.

Alternate Method:
Points E and F have same x co-ordinates i.e. 2.
Points E and F lie on a line parallel to Y-axis.
∴ The slope of EF is not defined.

iv. Here, G = (7, 1) and H = (-3, 1)
Slope of line GH = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{1-1}{-3-7}\) = o

Alternate Method:
Points G and H have same y co-ordinate i.e. 1.
∴ Points G and H lie on a line parallel to the
X-axis.
∴ The slope of GH is 0.

Question 2.
If the x and x-intercepts of line L are 2 and 3 respectively, then find the slope of line L.
Solution:
Given, x-intercept of line L is 2 and y-intercept of line L is 3
∴ The line L intersects X-axis at (2, 0) and Y-axis at (0,3).
∴ The line L passes through (2, 0) and (0, 3).
Slope of line L = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-0}{0-2}=\frac{-3}{2}\)

Question 3.
Find the slope of the line whose inclination is 30°.
Solution:
Given, inclination (θ) = 30°
∴ Slope of the line = tanθ = tan30° = \(\frac{1}{\sqrt{3}}\)

Question 4.
Find the slope of the line whose inclination is \(\frac{\pi}{4}\)
Solution:
Given, inclination (0) = \(\frac{\pi}{4}\)
∴ Slope of the line = tan θ = tan\(\frac{\pi}{4}\) = 1

Question 5.
A line makes intercepts 3 and 3 on the co-ordinate axes. Find the inclination of the line.
Solution:
Given, x-intercept of line is 3 and y-intercept of line is 3
∴ The line intersects X-axis at (3, 0) and Y-axis at (0, 3).
∴ The line passes through (3, 0) and (0,3).
∴ Slope of line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-0}{0-3}\) = -1
But, slope of a line = tan θ
∴ tan θ = – 1
= – tan \(\frac{\pi}{4}\)
= tan(π-\(\frac{\pi}{4}\) ) …[v tan(π – θ) = -tan θ]
tan θ = tan \(\frac{3\pi}{4}\)
θ = \(\frac{3\pi}{4}\)
The inclination of the line is \(\frac{3\pi}{4}\).
[Note: Answer given in the textbook is ‘-1 However, as per our calculation it is \(\frac{3\pi}{4}\)]

Question 6.
Without using Pythagoras theorem, show that points A (4, 4), B (3, 5) and C (- 1, – 1) are the vertices of a right-angled triangle.
Solution:
Given, A(4,4), B(3, 5), C (-1, -1).
Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-4}{3-4}\) = – 1
Slope of BC = \(\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}\)
Slope of AC = \(\frac{-1-4}{-1-4}\) = 1
Slope of AB x slope of AC = – 1 x 1 = – 1
∴ side AB ⊥ side AC
∴ ∆ABC is a right angled triangle right angled at A.
∴ The given points are the vertices of a right angled triangle.

Question 7.
Find the slope of the line which makes angle of 45° with the positive direction of the Y-axis measured anticlockwise.
Solution:

Since the line makes an angle of 45° with positive direction of Y-axis in anticlockwise direction,
Inclination of the line (0) = (90° + 45°)
∴ Slope of the line = tan(90° + 45°)
= – cot 45°
= -1

Question 8.
Find the value of k for which the points P(k, -1), Q(2,1) and R(4,5) are collinear.
Solution:
Given, points P(k, – 1), Q (2, 1) and R(4, 5) are collinear.
∴ Slope of PQ = Slope of QR .
∴ \(\frac{1-(-1)}{2-k}=\frac{5-1}{4-2}\)
∴ \(\frac{2}{2-k}=\frac{4}{2}\)
∴ 4 = 4 (2 – k)
∴ 1 = 2 – k
∴ k = 2 – 1 = 1

Question 9.
Find the acute angle between the X-axis and the line joining the points A(3, -1) and B(4, – 2).
Solution:
Given, A (3, – 1) and B (4, – 2)
Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-(-1)}{4-3}\) = – 1
But, slope of a line = tan θ
∴ tan θ = – 1
= – tan 45°
= tan (180° -45°)
… [∵ tan (180° – θ) = -tan θ]
= tan 135°
∴ θ = 135°

Let α be the acute angle that line AB makes with X-axis.
Then, α + 0 = 180°
α = 180°- 135° = 45°
∴ The acute angle between the X-axis and the line joining the points A and B is 45°.

Question 10.
A line passes through points A(xi, y0 and B(h, k). If the slope of the line is m, then show that k – y₁ = m (h – x₁).
Solution:
Given, A(x₁, y₁), B(h, k) and
slope of line AB = m
Slope of line AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
∴ m = \(\frac{\mathrm{k}-y_{1}}{\mathrm{~h}-x_{1}}\)
∴ k – y₁ = m (h – x₁)

Question 11.
If the points A(h, 0), B(0, k) and C(a, b) lie on a line, then show that \(\frac{a}{h}+\frac{b}{k}\) = 1. ‘
Solution:
Given, A(h, 0), B(0, k) and C(a, b)
Since the points A, B and C lie on a line, they are collinear.
∴ Slope of AB = slope of BC

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.1

Question 1.
If A(1, 3) and B(2, 1) are points, find the equation of the locus of point P such that PA = PB.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(1, 3), B(2, 1) and
PA = PB
∴ PA² = PB²
∴ (x – 1)² + ( y – 3)² = (x – 2)² + (y – 1)²
∴ x² – 2x + 1 + y² – 6y + 9 = x² – 4x + 4 + y² – 2y + 1
-2x – 6y + 10 = -4x – 2y + 5
∴ 2x – 4y + 5 = 0
∴ The required equation of locus is 2x – 4y + 5 = 0.

Question 2.
A(- 5,2) and B(4,1). Find the equation of the locus of point P, which is equidistant from A and B.
Solution:
Let P(x, y) be any point on the required locus.
P is equidistant from A(- 5, 2) and B(4, 1).
∴ PA = PB
∴ PA² = PB²
∴ (x + 5)² + (y – 2)² = (x – 4)² + (y – 1)²
∴ x² + 10x + 25 + y² — 4y + 4
= x² – 8x + 16 + y² – 2y + 1
∴ 10x – 4y + 29 = -8x – 2y + 17
∴ 18x – 2y + 12 = 0
∴ 9x – y + 6 = 0
The required equation of locus is 9x -y + 6 = 0.

Question 3.
If A(2, 0) and B(0, 3) are two points, find the equation of the locus of point P such that AP = 2BP.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(2, 0), B(0, 3) and AP = 2BP
∴ AP² = 4BP²
∴ (x – 2)² + (y – 0)² = 4[(x – 0)² + (y – 3)²]
∴ x² – 4x + 4 + y² = 4(x² + y² – 6y + 9)
x² – 4x + 4 + y² = 4x² + 4y² – 24y + 36
∴ 3x² + 3 y² + 4x – 24y + 32 = 0
∴ The required equation of locus is
3x² + 3y² + 4x – 24y + 32 = 0.
[Note: Answer given in the textbook , is
‘3x² + 3y² + 4x + 24y + 32 = O’.
However, as per our calculation it is ‘3x² + 3y² + 4x – 24y + 32 = 0’.]

Question 4.
If A(4,1) and B(5,4), find the equation of the locus of point P such that PA² = 3PB².
Solution:
Let P(x, y) be any point on the required locus. Given, A(4,1), B(5,4) and PA2 = 3PB2
∴ (x – 4)2 + (y – 1)2 = 3[(x – 5)² + (y – 4)²]
∴ x² – 8x + 16 + y² – 2y + 1 = 3(x² – 10x + 25 + y² – 8y + 16)
∴ x² – 8x + y² – 2y + 17 = 3x² -30x + 75 + 3y² – 24y + 48
∴ 2x² + 2y² – 22x – 22y + 106 = 0
∴ x² + y² – 11x – 11y + 53 = 0
∴ The required equation of locus is
x² + y² – 11x – 11y + 53 = 0.

Question 5.
A(2, 4) and B(5, 8), find the equation of the
locus of point P such that PA² – PB² = 13.
Solution:
Let P(x, y) be any point on the required locus. Given, A(2,4), B(5, 8) and PA²-PB² = 13
∴ [(x -2)² + (y – 4)²] – [(x -5)² + (y- 8)²] = 13
∴ (x² – 4x + 4 + y² – 8y + 16) – (x² – 10x + 25 + y² – 16y + 64) =13
∴ x² – 4x+ y² – 8y + 20 – x² + 10x – y² + 16y – 89 = 13
∴ 6x + 8y- 69 = 13
∴ 6x + 8y – 82 = 0
∴ 3x + 4y – 41 = 0
∴ The required equation of locus is 3x + 4y- 41 = 0.

Question 6.
A(1, 6) and B(3, 5), find the equation of the locus of point P such that segment AB subtends right angle at P. (∠APB = 90°)
Solution:
Let P(x, y) be any point on the required locus. Given,
A(l, 6) and B(3, 5),
∠APB = 90°
∴ ΔAPB is a right angled triangle,
By Pythagoras theorem,

AP² + PB² = AB² P (x,y)
∴ [(x – 1)² + (y – 6)²] + [(x – 3)² + (y – 5)²] = (1 – 3)² + (6 -5)²
∴ x² — 2x + 1 + y² — 12y + 36 + x² – 6x + 9 + y² – 10y + 25 = 4 + 1
∴ 2x² + 2y² – 8x – 22y + 66 = 0
∴ x² + y² – 4x – 11y + 33 = 0
∴ The required equation of locus is x² + y² – 4x – 11y + 33 = 0.
[Note: Answer given in the textbook is
‘3x² + 4y² – 4x – 11y + 33 = 0’.
However, as per our calculation it is ‘x² + y² – 4x – 11y + 33 = O

Question 7.
If the origin is shifted to the point 0′(2, 3), the axes remaining parallel to the original axes, find the new co-ordinates of the points
i. A(1, 3) ii. B(2,5)
Solution:
Origin is shifted to (2, 3) = (h, k)
Let the new co-ordinates be (X, Y).
x = X + handy = Y + k
x = X + 2 andy = Y + 3 …(i)

i. Given, A(x, y) = A( 1, 3)
x = X + 2 andy = Y + 3 …[From(i)]
∴ 1 = X + 2 and 3 = Y + 3 X = – 1 and Y = 0
∴ The new co-ordinates of point A are (- 1,0).

ii. Given, B(x, y) = B(2, 5)
x = X + 2 and y = Y + 3 …[From(i)]
∴ 2 = X + 2 and 5 = Y + 3
∴ X = 0 and Y = 2
∴ The new co-ordinates of point B are (0, 2).

Question 8.
If the origin is shifted to the point O'(1, 3), the axes remaining parallel to the original axes, find the old co-ordinates of the points
i. C(5,4) ii. D(3,3)
Solution:
Origin is shifted to (1,3) = (h, k)
Let the new co-ordinates be (X, Y).
x = X + h andy = Y + k
∴ x = X+1 andy = Y + 3 …(i)

i. Given, C(X, Y) = C(5, 4)
x = X +1 andy = Y + 3 …[From(i)]
∴ x = 5 + 1 = 6 andy = 4 + 3 = 7
∴ The old co-ordinates of point C are (6, 7).

ii. Given, D(X, Y) = D(3, 3)
x = X + 1 andy = Y + 3 …[From(i)]
∴ x = 3 + 1 = 4 and y = 3 + 3 = 6
∴ The old co-ordinates of point D are (4, 6).

Question 9.
If the co-ordinates A(5, 14) change to B(8, 3) by shift of origin, find the co-ordinates of the point, where the origin is shifted.
Solution:
Let the origin be shifted to (h, k).
Given, A(x, y) = A(5,14), B(X, Y) = B(8, 3)
Since x = X + h andy = Y + k,
5 = 8 + hand 14 = 3 + k ,
∴ h = – 3 and k = 11
The co-ordinates of the point, where the origin is shifted are (- 3, 11).

Question 10.
Obtain the new equations of the following loci if the origin is shifted to the point 0′(2,2), the direction of axes remaining the same:
i. 3x-y + 2 = 0
11. x²+y²-3x = 7
iii. xy – 2x – 2y + 4 = 0
iv. y² – 4x – 4y + 12 = 0
Solution:
Given, (h,k) = (2,2)
Let (X, Y) be the new co-ordinates of the point (x,y).
∴ x = X + handy = Y + k
∴ x = X + 2 andy = Y + 2
i. Substituting the values of x and y in the equation 3x -y + 2 = 0, we get
3(X + 2) – (Y + 2) + 2 = 0
∴ 3X + 6-Y-2 + 2 = 0
∴ 3 X – Y + 6 = 0, which is the new equation of locus.

ii. Substituting the values of x and y in the equation
x² + y² – 3x = 7, we get
(X + 2)² + (Y + 2)² – 3(X + 2) = 7
∴ X² + 4X + 4 + Y² + 4Y + 4 – 3X – 6 = 7
∴ X² + Y² + X + 4Y – 5 = 0, which is the new
equation of locus.

iii; Substituting the values of x and y in the equation xy – 2x – 2y + 4 = 0, we get
(X + 2) (Y + 2) – 2(X + 2) – 2(Y + 2) + 4 = 0
∴ XY + 2X + 2Y + 4 – 2X – 4-2Y- 4 + 4 = 0
∴ XY = 0, which is the new equation of locus.

iv. Substituting the values of x and y in the equation y² – 4x – 4y + 12 = 0, we get
(Y + 2)² – 4(X + 2) – 4(Y + 2) + 12 = 0
∴ Y² + 4Y + 4 – 4X – 8 – 4Y -8 + 12 = 0
∴ Y² – 4X = 0, which is the new equation of locus.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

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

Question 1.
Given A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 2
\end{array}\right]\), I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) if A – λI is a singular matrix, then ___________
(a) λ = 0
(b) λ² – 3λ – 4 = 0
(c) λ² + 3λ – 4 = 0
(d) λ² – 3λ – 6 = 0
Answer:
(b) λ² – 3λ – 4 = 0
Hint:

Question 2.
Consider the matrices A = \(\left[\begin{array}{ccc}
4 & 6 & -1 \\
3 & 0 & 2 \\
1 & -2 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 4 \\
0 & 1 \\
-1 & 2
\end{array}\right]\), C = \(\left[\begin{array}{l}
3 \\
1 \\
2
\end{array}\right]\). Out of the given matrix products, ___________
(i) (AB)^TC
(ii) C^TC(AB)^T
(iii) C^TAB
(iv) A^TABB^TC
(a) Exactly one is defined
(b) Exactly two are defined
(c) Exactly three are defined
(d) all four are defined
Answer:
(c) Exactly three are defined
Hint:
A is of order 3 × 3, B is of order 3 × 2 and C is of order 3 × 1.
(AB)^TC is of order 2 × 1.
C^TC and (AB)^T are of different orders.
C^TC (AB)^T is not defined.
C^TAB is of order 1 × 2.
A^TABB^TC is of order 3 × 1.

Question 3.
If A and B are square matrices of equal order, then which one is correct among the following?
(a) A + B = B + A
(b) A + B = A – B
(c) A – B = B – A
(d) AB = BA
Answer:
(a) A + B = B + A
Hint:
Matrix addition is commutative.
∴ A + B = B + A

Question 4.
If A = \(\left[\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
a & 2 & b
\end{array}\right]\) is a matrix satisfying the equation AA^T = 9I, where I is the identity matrix of order 3, then the ordered pair (a, b) is equal to ___________
(a) (2, -1)
(b) (-2, 1)
(c) (2, 1)
(d) (-2, -1)
Answer:
(d) (-2, -1)

Question 5.
If A = \(\left[\begin{array}{ll}
\alpha & 2 \\
2 & \alpha
\end{array}\right]\) and |A³| = 125, then α = ___________
(a) ±3
(b) ±2
(c) ±5
(d) 0
Answer:
(a) ±3
Hint:
|A³| = 125
|A|³ = 53 …….[∵ |Aⁿ| = |A|ⁿ, n ∈ N]
∴ |A| = 5
\(\left|\begin{array}{ll}
\alpha & 2 \\
2 & \alpha
\end{array}\right|=5\)
α² – 4 = 5
α² = 9
∴ α = ± 3

Question 6.
If \(\left[\begin{array}{ll}
5 & 7 \\
x & 1 \\
2 & 6
\end{array}\right]-\left[\begin{array}{cc}
1 & 2 \\
-3 & 5 \\
2 & y
\end{array}\right]=\left[\begin{array}{cc}
4 & 5 \\
4 & -4 \\
0 & 4
\end{array}\right]\), then ___________
(a) x = 1, y = -2
(b) x = -1, y = 2
(c) x = 1, y = 2
(d) x = -1, y = -2
Answer:
(c) x = 1, y = 2

Question 7.
If A + B = \(\left[\begin{array}{ll}
7 & 4 \\
8 & 9
\end{array}\right]\) and A – B = \(\left[\begin{array}{ll}
1 & 2 \\
0 & 3
\end{array}\right]\), then the value of A is ___________
(a) \(\left[\begin{array}{ll}
3 & 1 \\
4 & 3
\end{array}\right]\)
(b) \(\left[\begin{array}{ll}
4 & 3 \\
4 & 6
\end{array}\right]\)
(c) \(\left[\begin{array}{ll}
6 & 2 \\
8 & 6
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
7 & 6 \\
8 & 12
\end{array}\right]\)
Answer:
(b) \(\left[\begin{array}{ll}
4 & 3 \\
4 & 6
\end{array}\right]\)

Question 8.
If \(\left[\begin{array}{cc}
x & 3 x-y \\
z x+z & 3 y-w
\end{array}\right]=\left[\begin{array}{ll}
3 & 2 \\
4 & 7
\end{array}\right]\), then ___________
(a) x = 3, y = 7, z = 1, w = 14
(b) x = 3, y = -5, z = -1, w = -4
(c) x = 3, y = 6, z = 2, w = 7
(d) x = -3, y = -7, z = -1, w = -14
Answer:
(a) x = 3, y = 7, z = 1, w = 14

Question 9.
For suitable matrices A, B, the false statement is ___________
(a) (AB)^T = A^TB^T
(B) (A^T)^T = A
(C) (A – B)^T = A^T – B^T
(D) (A + B)^T = A^T + B^T
Answer:
(a) (AB)^T = A^TB^T
Hint:
(AB)^T = B^TA^T

Question 10.
If A = \(\left[\begin{array}{cc}
-2 & 1 \\
0 & 3
\end{array}\right]\) and f(x) = 2x² – 3x, then f(A) = ___________
(a) \(\left[\begin{array}{cc}
14 & 1 \\
0 & -9
\end{array}\right]\)
(b) \(\left[\begin{array}{cc}
-14 & 1 \\
0 & 9
\end{array}\right]\)
(c) \(\left[\begin{array}{cc}
14 & -1 \\
0 & 9
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
-14 & -1 \\
0 & -9
\end{array}\right]\)
Answer:
(c) \(\left[\begin{array}{cc}
14 & -1 \\
0 & 9
\end{array}\right]\)
Hint:

(II) Answer the following questions.

Question 1.
If A = diag[2, -3, -5], B = diag[4, -6, -3] and C = diag [-3, 4, 1], then find
i. B + C – A
ii. 2A + B – 5C.
Solution:

Question 2.
If f(α) = A = \(\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\), find
i. f(-α)
ii. f(-α) + f(α)
Solution:

Question 3.
Find matrices A and B, where
(i) 2A – B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\) and A + 3B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\)
(ii) 3A – B = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 0 & 5
\end{array}\right]\) and A + 5B = \(\left[\begin{array}{ccc}
0 & 0 & 1 \\
-1 & 0 & 0
\end{array}\right]\)
Solution:
i. Given equations are
2A – B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\) …….(i)

Question 4.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
3 & -2 \\
-1 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
-3 & 4 & 1 \\
2 & -1 & -3
\end{array}\right]\), verify
i. (A + 2B^T)^T = A^T + 2B
ii. (3A – 5B^T)^T = 3A^T – 5B.
Solution:

Question 5.
If A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) and A + A^T = I, where I is a unit matrix of order 2 × 2, then find the value of α.
Solution:

Question 6.
If A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & 2 \\
-1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & 3 & 2 \\
4 & -1 & -3
\end{array}\right]\), show that AB is singular.
Solution:

Question 7.
If A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 4 & 6 \\
1 & 2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
-3 & 2 & -1 \\
-2 & 1 & 0
\end{array}\right]\), show that AB and BA are both singular matrices.
Solution:

Question 8.
If A = \(\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
2 & -1 & 5
\end{array}\right]\), show that BA = 6I.
Solution:

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

Question 10.
If Aα = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), show that Aα . Aβ = Aα+β
Solution:

Question 11.
If A = \(\left[\begin{array}{cc}
1 & \omega \\
\omega^{2} & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
\omega^{2} & 1 \\
1 & \omega
\end{array}\right]\), where ω is a complex cube root of unity, then show that AB + BA + A – 2B is a null matrix.
Solution:
ω is the complex cube root of unity.
ω³ = 1
ω³ – 1 = 0
(ω – 1) (ω² + ω + 1) = 0
ω = 1 or ω² + ω + 1 = 0
But, ω is a complex number.
1 + ω + ω² = 0 …….(i)
AB + BA + A – 2B

which is a null matrix.

Question 12.
If A = \(\left[\begin{array}{lrr}
2 & -2 & -4 \\
-1 & 3 & 4 \\
1 & -2 & -3
\end{array}\right]\), show that A² = A.
Solution:

Question 13.
If A = \(\left[\begin{array}{ccc}
4 & -1 & -4 \\
3 & 0 & -4 \\
3 & -1 & -3
\end{array}\right]\), show that A² = I.
Solution:

Question 14.
If A = \(\left[\begin{array}{cc}
3 & -5 \\
-4 & 2
\end{array}\right]\), show that A² – 5A – 14I = 0.
Solution:

Question 15.
If A = \(\left[\begin{array}{cc}
2 & -1 \\
-1 & 2
\end{array}\right]\), show that A – 4A + 3I = 0.
Solution:

Question 16.
If A = \(\left[\begin{array}{cc}
-3 & 2 \\
2 & -4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & x \\
y & 0
\end{array}\right]\) and (A + B)(A – B) = A² – B², find x and y.
Solution:
(A + B)(A – B) = A² – B²
A² – AB + BA – B² = A² – B²
-AB + BA = 0
AB = BA

By equality of matrices, we get
2 – 4x = -3x
∴ x = 2 and 2y = 2x
y = x
∴ y = 2
∴ x = 2, y = 2

Question 17.
If A = \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]\), show that (A + B)(A – B) ≠ A² – B².
Solution:
We have to prove that
(A + B) . (A – B) ≠ A² – B²
i.e., to prove that A(A – B) + B(A – B) ≠ A² – B²
i.e., to prove that A² – AB + BA – B² ≠ A² – B²
i.e., to prove that AB ≠ BA.

∴ AB ≠ BA

Question 18.
If A = \(\left[\begin{array}{ll}
2 & -1 \\
3 & -2
\end{array}\right]\), find A³.
Solution:

∴ A² = I
Multiplying throughout by A, we get
A³ = A . I
∴ A³ = A

Question 19.
Find x, y if,

Solution:

Question 20.
Find x, y, z if

Solution:

Question 21.
If A = \(\left[\begin{array}{ccc}
2 & 1 & -3 \\
0 & 2 & 6
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & 0 & -2 \\
3 & -1 & 4
\end{array}\right]\), find AB^T and A^TB.
Solution:

Question 22.
If A = \(\left[\begin{array}{cc}
2 & -4 \\
3 & -2 \\
0 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 1 & 0
\end{array}\right]\), show that (AB)^T = B^TA^T.
Solution:

Question 23.
If A = \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\), prove that Aⁿ = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\), for all n ∈ N.
Solution:

Question 24.
If A = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), prove that Aⁿ = \(\left[\begin{array}{cc}
\cos n \theta & \sin n \theta \\
-\sin n \theta & \cos n \theta
\end{array}\right]\), for all n ∈ N.
Solution:

Question 25.
Two farmers Shantaram and Kantaram cultivate three crops rice, wheat, and groundnut. The sale (in Rupees) of these crops by both the farmers for the month of April and may 2008 is given below,

Find
(i) the total sale in rupees for two months of each farmer for each crop.
(ii) the increase in sales from April to May for every crop of each farmer.
Solution:
(i) Total sale for Shantaram:
For rice = 15000 + 18000 = ₹ 33000.
For wheat = 13000 + 15000 = ₹ 28000.
For groundnut = 12000 + 12000 = ₹ 24000.
Total sale for Kantaram:
For rice = 18000 + 21000 = ₹ 39000
For wheat = 15000 + 16500 = ₹ 31500
For groundnut = 8000 + 16000 = ₹ 24000

Alternate method:
Matrix form

∴ The total sale of April and May of Shantaram in ₹ is ₹ 33000 (rice), ₹ 28000 (wheat), ₹ 24000 (groundnut), and that of Kantaram in ₹ is ₹ 39000(rice), ₹ 31500(wheat), and ₹ 24000 (groundnut).

(ii) Increase in sale from April to May for Shantaram:
For rice = 18000 – 15000 = ₹ 3000
For wheat = 15000 – 13000 = ₹ 2000
For groundnut = 12000 – 12000 = ₹ 0
Increase in sale from April to May for Kantaram:
For rice = 21000 – 18000 = ₹ 3000
For wheat = 16500 – 15000 = ₹ 1500
For groundnut = 16000 – 8000 = ₹ 8000

Alternate method:
Matrix form

∴ The increase in sales for Shantaram from April to May in each crop is ₹ 3000 (rice), ₹ 2000(wheat), 0 (groundnut), and that for Kantaram is ₹ 3000 (rice), ₹ 1500 (wheat), and ₹ 8000 (groundnut).