ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

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ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

Very short answer/objective questions (1 to 3) :

Question 1.
Find the values of :
(i) cos 210°
(ii) sin 225°
(iii) tan 330°
(iv) sin 930°
(v) cos (- 870°)
(vi) cos \(\frac{2 \pi}{3}\)
(vii) sin \(\frac{19 \pi}{4}\)
(viii) tan (- \(\frac{4 \pi}{3}\))
Solution:
(i) cos 210° = cos (180° + 30°)
= – cos 30°
= – \(\frac{\sqrt{3}}{2}\)
[∵ cos (180° + θ) = – cos θ]

(ii) sin 225° = sin (180° + 45°)
= – sin 45°
= – \(\frac{1}{\sqrt{2}}\)
[∵ sin (180° + θ) = – sin θ]

(iii) tan 330° = tan (360° – 30°)
= – tan 30°
= – \(\frac{1}{\sqrt{3}}\)
[∵ tan (360° – θ) = – tan θ]

(iv) sin 930° = sin (720° + 210°)
= sin 210°
= sin (180° + 30°)
= – sin 30°
= – \(\frac{1}{2}\)
[∵ sin (180° + θ) = – sin θ]

(v) cos (- 870°) = cos 870°
[∵ cos (- θ) = cos θ]
= cos (720° + 150°)
= cos 150°
= cos (180° – 30°)
= – cos 30°
= – \(\)
[∵ cos (180° – θ) = – cos θ]

(vi) cos \(\frac{2 \pi}{3}\) = cos (π – \(\frac{\pi}{3}\))
= – cos \(\frac{\pi}{3}\)
= – \(\frac{1}{2}\)

(vii) sin \(\frac{19 \pi}{4}\) = sin (4π + \(\frac{3 \pi}{4}\))
= sin \(\frac{3 \pi}{4}\)
= sin (π – \(\frac{\pi}{4}\))
= sin \(\frac{\pi}{4}\)
= \(\frac{1}{\sqrt{2}}\)
[∵ sin (π – θ) = sin θ]

(viii) tan (- \(\frac{4 \pi}{3}\)) = – tan (\(\frac{4 \pi}{3}\))
[∵ tan (- θ) = – tan θ]
= – taan (π + \(\frac{\pi}{3}\))
= – tan \(\frac{\pi}{3}\)
= – √3
[∵ tan (π + θ) = tan θ]

Question 2.
Find the values of :
(i) sin² \(\frac{\pi}{2}\) + cos² \(\frac{\pi}{3}\) – tan² \(\frac{\pi}{4}\)
(ii) tan² \(\frac{\pi}{3}\) + 2 cos² \(\frac{\pi}{4}\) + 3 sec² \(\frac{\pi}{6}\) – 4 cos² \(\frac{\pi}{2}\).
Solution:
(i) sin² \(\frac{\pi}{2}\) + cos² \(\frac{\pi}{3}\) – tan² \(\frac{\pi}{4}\)
= \(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2\) – 1²
= \(\frac{1}{4}+\frac{1}{4}\) – 1 = – \(\frac{1}{2}\)

(ii) tan² \(\frac{\pi}{3}\) + 2 cos² \(\frac{\pi}{4}\) + 3 sec² \(\frac{\pi}{6}\) – 4 cos² \(\frac{\pi}{2}\)
= (√3)² + 2 \(\left(\frac{1}{\sqrt{2}}\right)^2\) + 3 \(\left(\frac{2}{\sqrt{3}}\right)^2\) – 4 × 0²
= 3 + 2 × \(\frac{1}{2}\) + 3 × \(\frac{4}{3}\) – 0
= 3 + 1 + 4 = 8

Question 3.
If tan x = \(\frac{2}{3}\) and tan y = \(\frac{3}{4}\), find the value of tan (x + y).
Solution:
Given tan x = \(\frac{2}{3}\) ;
tan y = \(\frac{3}{4}\)
∴ tan (x + y) = \(\frac{\tan x+\tan y}{1-\tan x \tan y}\)
= \(\frac{\frac{2}{3}+\frac{3}{4}}{1-\frac{2}{3} \times \frac{3}{4}}\)
= \(\frac{\frac{17}{12}}{\frac{1}{2}}\)
= \(\frac{34}{12}=\frac{17}{6}\)

Short answer questions (4 to 6) :

Question 4.
Express the following as functions of angles less than 45° :
(i) sin (- 1785°)
(ii) cosec (- 7498°)
Soluion:
(i) sin (- 1785°) = – sin (1 785°)
[∵ sin(- θ) = – sin θ]
= – sin (1440° + 345°)
= – sin (4 X 360° + 3450)
= – sin (345°)
= – sin (360 – 15°)
= sin 15°
[∵ sin (360° – θ) = – sin θ]
= sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
[∵ sin (A – B) = sin A cos B – cos A sin B]
= \(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2}\)
= \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)

(ii) cosec (- 7498°) = – cosec (7498°)
[∵ cosec (- θ) = – cosec θ]
= – cosec (360° × 20 + 298°)
– cosec (298°)
= – cosec (270° + 28°)
= sec 28°
[∵ cosec (270° + θ) – sec θ]

Question 5.
(i) Which is bigger: sin 55° or cos 55°?
(ii) If θ = 100°, determine the sign of (sin θ + cos θ).
Solution:
Now cos 55° = cos (90° – 35°) = sin 35°
since sin x be an increasing function in first quadrant
since 55° > 35°
⇒ sin 55° > sin 35° = cos 55°
[if x₁ > x₂ and f be an increasing function.
Then f (x₁) > f (x₂)]
Thus sin 55° is bigger.

(ii) When θ = 1000
∴ sin θ + cos θ = sn 100° + cos 100°
= sin 100° + cos (90° + 10°)
= sin 100°- sin 10°
[∵ cos (90° + θ) = – sin θ]
= sin (90° + 10°) – sin 10°
= cos 10° – sin 10° > 0
[∵ When 0 < θ < \(\frac{\pi}{4}\) ⇒ cos θ > sin θ]

Question 6.
Prove that :
(i) cot A + tan (π + A) + tan (\(\frac{\pi}{2}\) + A) + tan (2π – A) = 0
(ii) \(2 \sin ^2 \frac{\pi}{6}+\ {cosec} \frac{7 \pi}{6} \cos ^2 \frac{\pi}{3}\) = 0
(iii) \(3 \cos ^2 \frac{\pi}{4}+\sec \frac{2 \pi}{3}+5 \tan ^2 \frac{\pi}{3}=\frac{29}{2}\)
(iv) \(2 \cos ^2 \frac{\pi}{4}+2 \sin ^2 \frac{3 \pi}{4}+2 \sec ^2 \frac{\pi}{3}\) = 10
(v) \(\cot ^2 \frac{\pi}{6}+\ {cosec} \frac{5 \pi}{6}+3 \tan ^2 \frac{\pi}{6}\) = 6
(vi) \(\left(3 \cos \frac{\pi}{3} \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \tan \frac{\pi}{4}\right)\) cos 2π = 1
Solution:
(i) L.H.S = cot A + tan (π + A) + tan (\(\frac{\pi}{2}\) + A) + tan (2π – A)
= cot A + tan A – cot A – tan A = 0
= R.H.S.

(ii) L.H.S. = \(2 \sin ^2 \frac{\pi}{6}+\ {cosec} \frac{7 \pi}{6} \cos ^2 \frac{\pi}{3}\)
= 2 (\(\frac{1}{2}\))² + cosec (π + \(\frac{\pi}{6}\)) × (\(\frac{1}{2}\))²
= 2 × \(\frac{1}{4}\) – cosec \(\frac{\pi}{6} \times \frac{1}{4}\)
[∵ cosec (π + θ) = – cosec θ]
= \(\frac{1}{2}-2 \times \frac{1}{4}\) = 0
= R.H.S.

(iii) L.H.S. = \(3 \cos ^2 \frac{\pi}{4}+\sec \frac{2 \pi}{3}+5 \tan ^2 \frac{\pi}{3}\)
= \(3\left(\frac{1}{\sqrt{2}}\right)^2+\sec \left(\pi-\frac{\pi}{3}\right)+5(\sqrt{3})^2\)
= \(\frac{3}{2}-\sec \frac{\pi}{3}\) + 15
[∵ sec (π – θ) = – cosec θ]
= \(\frac{3}{2}\) – 2 + 15
= \(\frac{29}{2}\)
= R.H.S.

(iv) L.H.S. = \(2 \cos ^2 \frac{\pi}{4}+2 \sin ^2 \frac{3 \pi}{4}+2 \sec ^2 \frac{\pi}{3}\)
= \(2\left(\frac{1}{\sqrt{2}}\right)^2+2 \sin ^2\left(\pi-\frac{\pi}{4}\right)\) + 2 (2)²
= 2 × \(\frac{1}{2}\) + 2 \(\left(\frac{1}{\sqrt{2}}\right)^2\) + 8
= 1 + 1 + 8 = 10
= R.H.S.

(v) L.H.S. = \(\cot ^2 \frac{\pi}{6}+\ {cosec} \frac{5 \pi}{6}+3 \tan ^2 \frac{\pi}{6}\)
= \((\sqrt{3})^2+\ {cosec}\left(\pi-\frac{\pi}{6}\right)+3\left(\frac{1}{\sqrt{3}}\right)^2\)
= 3 + cosec \(\frac{\pi}{6}\) + 3 × \(\frac{1}{3}\)
[∵ cosec (π – θ) = cosec θ]
= 3 + 2 + 1 = 6
= R.H.S.

(vi) L.H.S. = \(\left(3 \cos \frac{\pi}{3} \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \tan \frac{\pi}{4}\right)\) cos 2π
= [3 × \(\frac{1}{2}\) × 2 – 4 sin (π – \(\frac{\pi}{6}\)) × 1] × 1
= [3 – 4 sin \(\frac{\pi}{6}\)]
= 3 – 4 × \(\frac{1}{2}\)
= 3 – 2 = 1
= R.H.S.

Long answer questions (7 to 32) :

Question 7.
Evaluate the following :
(i) √2 sin 135° cos 210° tan 240° cot 300° sec 330°
(ii) sin 690° cos 930° + tan (- 765°) cosec (- 1170°)
(iii) \(\sin ^2 \frac{\pi}{8}+\sin ^2 \frac{3 \pi}{8}+\sin ^2 \frac{5 \pi}{8}+\sin ^2 \frac{7 \pi}{8}\)
(iv) \(\sin ^2 \frac{\pi}{4}+\sin ^2 \frac{3 \pi}{4}+\sin ^2 \frac{5 \pi}{4}+\sin ^2 \frac{7 \pi}{4}\)
(v) \(\tan \frac{\pi}{12} \tan \frac{\pi}{16} \tan \frac{5 \pi}{12} \tan \frac{7 \pi}{16}\)
Solution:
(i) sin 135° = sin(180° – 45°)
= sin 45°
= \(\frac{1}{\sqrt{2}}\)
[∵ sin(180° – θ) = sin θ]
cos 210° = cos(180° + 30°)
= – cos 30°
= – \(\frac{\sqrt{3}}{2}\)
tan 240° = tan (180° + 60°)
= tan 60° = √3
cot 300° = cot(360° – 60°)
= – cot 60°
= \(\frac{1}{\sqrt{32}}\)
sec 330° = sec (360° – 30°)
= sec 30°
= \(\frac{2}{\sqrt{3}}\)
Thus, √2 sin 135° cos 210° tan 240° cot 300° sec 330° = \(\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\left(-\frac{\sqrt{3}}{2}\right) \sqrt{3}\left(-\frac{1}{\sqrt{3}}\right)\left(\frac{2}{\sqrt{3}}\right)\) = 1

(ii) Now sin 690° = sin (720° – 30°)
= sin (- 30°)
= – \(\frac{1}{2}\)
cos 930° = cos (720° + 210°)
= cos 210°
= cos (180° + 30°)
= – cos30°
tan (- 765°) = – tan (765°)
= – tan (720° + 45°)
= – tan 45° = – 1
and cosec (- 1170°) = – cosec (1080° + 90°)
= – cosec 90° – 1
Now, sin 690° cos 930° + tan (- 765°) cosec (- 1170°) = \(\left(-\frac{1}{2}\right)\left(-\frac{\sqrt{3}}{2}\right)\) + (- 1) (- 1)
= \(\frac{\sqrt{3}}{4}\) + 1

(iii) \(\sin ^2 \frac{\pi}{8}+\sin ^2 \frac{3 \pi}{8}+\sin ^2 \frac{5 \pi}{8}+\sin ^2 \frac{7 \pi}{8}\)

(iv) \(\sin ^2 \frac{\pi}{4}+\sin ^2 \frac{3 \pi}{4}+\sin ^2 \frac{5 \pi}{4}+\sin ^2 \frac{7 \pi}{4}\)
= \(\sin ^2 \frac{\pi}{4}+\sin ^2\left(\frac{\pi}{2}+\frac{\pi}{4}\right)+\sin ^2 \frac{5 \pi}{4}+\sin ^2\left(\frac{\pi}{2}+\frac{5 \pi}{4}\right)\)
= \(\sin ^2 \frac{\pi}{4}+\cos ^2 \frac{\pi}{4}+\sin ^2 \frac{5 \pi}{4}+\cos ^2 \frac{5 \pi}{4}\)
[∵ sin (\(\frac{\pi}{2}\) + θ) = cos θ]
= 1 + 1 = 2

Aliter:

(v) \(\tan \frac{\pi}{12} \tan \frac{\pi}{16} \tan \frac{5 \pi}{12} \tan \frac{7 \pi}{16}\)

Question 8.
Prove that :
cos x + sin (\(\frac{3 \pi}{2}\) + x) – sin (\(\frac{3 \pi}{2}\) – x) + cos (π + x) = 0.
Solution:
L.H.S. = cos x + sin (\(\frac{3 \pi}{2}\) + x) – sin (\(\frac{3 \pi}{2}\) – x) + cos (π + x)
= cos x – cos x + cos x – cos x
= 0
= R.H.S.

Question 9.
Simplify the following:
(i) \(\frac{\cos x}{\sin \left(\frac{\pi}{2}+x\right)}+\frac{\sin (-x)}{\sin (\pi+x)}-\frac{\tan \left(\frac{\pi}{2}+x\right)}{\cot x}\)
Soluion:
\(\frac{\cos x}{\sin \left(\frac{\pi}{2}+x\right)}+\frac{\sin (-x)}{\sin (\pi+x)}-\frac{\tan \left(\frac{\pi}{2}+x\right)}{\cot x}\)
= \(\frac{\cos x}{\cos x}-\frac{\sin x}{-\sin x}-\frac{(-\cot x)}{\cot x}\)
= 1 + 1 + 1
= 3

(ii) \(\frac{\sin (\pi+x) \cos \left(\frac{\pi}{2}+x\right) \tan \left(\frac{3 \pi}{2}-x\right) \cot (2 \pi-x)}{\sin (2 \pi-x) \cos (2 \pi+x) \ {cosec}(-x) \sin \left(\frac{3 \pi}{2}+x\right)}\)
Soluion:
\(\frac{\sin (\pi+x) \cos \left(\frac{\pi}{2}+x\right) \tan \left(\frac{3 \pi}{2}-x\right) \cot (2 \pi-x)}{\sin (2 \pi-x) \cos (2 \pi+x) \ {cosec}(-x) \sin \left(\frac{3 \pi}{2}+x\right)}\)
= \(\frac{(-\sin x)(-\sin x)(\cot x)(-\cot x)}{(-\sin x) \cos x(-\ {cosec} x)(-\cos x)}\)
= \(\frac{\sin ^2 x \cot ^2 x}{\cos ^2 x}\)
= tan² x cot² x = 1

Question 10.
Find y from the following equation:
cosec (\(\frac{\pi}{2}\) + x) + y cos x cot (\(\frac{\pi}{2}\) + x) = sin (\(\frac{\pi}{2}\) + x)
Solution:
Given,
cosec (\(\frac{\pi}{2}\) + x) + y cos x cot (\(\frac{\pi}{2}\) + x) = sin (\(\frac{\pi}{2}\) + x)
⇒ + sec x + y cos x (- tan x) = cos x
⇒ sec x – cos x = y cos x \(\left(\frac{\sin x}{\cos x}\right)\)
⇒ \(\frac{1}{\ cos x}\) – cos x = y sin x
⇒ \(\frac{1-\cos ^2 x}{\cos x}\) = y sin x
⇒ \(\frac{\sin ^2 x}{\cos x}\) = y sin x
⇒ y = tan x

Question 11.
If 8x = π, show that cos 7x + cos x = 0.
Solution:
Given 8x = π
L.H.S. = cos 7x + cos x
= cos \(\frac{7 \pi}{8}\) + cos \(\frac{\pi}{8}\)
= cos (π – \(\frac{\pi}{8}\)) + cos \(\frac{\pi}{8}\)
= – cos \(\frac{\pi}{8}\) + cos \(\frac{\pi}{8}\)
= 0
= R.H.S.
= \(-\frac{1}{\tan 25^{\circ}}=-\frac{1}{x}\)
[∵ cos (π – θ) = – cos θ]

Question 12.
If tan 25° = x, prove that \(\frac{\tan 155^{\circ}-\tan 115^{\circ}}{1+\tan 155^{\circ} \tan 115^{\circ}}=\frac{1-x^2}{2 x}\)
Solution:
Now, tan 155° = tan (180° – 25°)
= – tan 25° = – x
[∵ tan 25° = x]
and tan 115° = tan(90° + 25°)
= – cot 25°
= \(-\frac{1}{\tan 25^{\circ}}=\frac{-1}{x}\)
LH.S. = \(\frac{\tan 155^{\circ}-\tan 115^{\circ}}{1+\tan 155^{\circ} \tan 115^{\circ}}\)
= \(\frac{-x-\left(-\frac{1}{x}\right)}{1+(-x)\left(-\frac{1}{x}\right)}\)
= \(\frac{-x+\frac{1}{x}}{1+1}\)
= \(\frac{1-x^2}{2 x}\)
= R.H.S.

Question 13.
Prove that \(\frac{\cos 135^{\circ}-\cos 120^{\circ}}{\cos 135^{\circ}+\cos 120^{\circ}}\) = 3 – 2√2
Solution:
L.H.S. = \(\frac{\cos 135^{\circ}-\cos 120^{\circ}}{\cos 135^{\circ}+\cos 120^{\circ}}\)

Question 14.
If A, B, C are angles of a triangle, prove that
(i) sin (A + B) = sin C
(ii) cos \(\frac{A+B}{2}\) = sin \(\frac{C}{2}\)
(iii) \(\frac{\tan (B+C)+\tan (C+A)+\tan (A+B)}{\tan (\pi-A)+\tan (2 \pi-B)+\tan (3 \pi-C)}\) = 1
Solution:
Since A, B, C are the angles of a triangle
∴ A + B + C = π
⇒ A + B = π – C

(i) sin (A + B) = sin (π – C)
= sin C

(ii) cos \(\frac{A+B}{2}\) = cos \(\left(\frac{\pi-C}{2}\right)\)
= cos \(\left(\frac{\pi}{2}-\frac{C}{2}\right)\)
= sin \(\frac{C}{2}\)

(iii) \(\frac{\tan (B+C)+\tan (C+A)+\tan (A+B)}{\tan (\pi-A)+\tan (2 \pi-B)+\tan (3 \pi-C)}\)

Question 15.
If A, B, C, D are angles of a quadrilateral, then prove that
(i) sin (A + D) + sin (B + C) = 0
(ii) \(\sin \left(\frac{A+B}{2}\right)=\sin \left(\frac{C+D}{2}\right)\)
(iii) \(\tan \left(\frac{A+B}{2}\right)+\tan \left(\frac{C+D}{2}\right)\) = 0
(iv) cosec² (A + B) = cosec² (C + D)
(v) sin A + sin (B + C + D ) = 0.
Solution:
(i) Since A, B, C and D are the angles of a quadrilateral.
∴ A + B + C + D = 360°
⇒ A + D = 360° – (B + C)
⇒ sin (A + D) = sin \(\left[360^{\circ}-\overline{\mathrm{B}+\mathrm{C}}\right]\)
= – sin (B + C)
sin (A + D) + sin (B + C) = 0

(ii) A + B + C + D = 360°

(iii) ∴ tan \(\left(\frac{A+B}{2}\right)\)
= tan [180° – \(\left(\frac{C+D}{2}\right)\)] [using (1)]
= tan \(\left(\frac{C+D}{2}\right)\)
[∵ tan (180 – θ) = – tan θ]
⇒ \(\tan \left(\frac{A+B}{2}\right)+\tan \left(\frac{C+D}{2}\right)\) = 0

(iv) since A + B + C + D = 360°
⇒ A + B = 360° – (C +D)
⇒ cosec (A + B) = cosec² [360° – (C + D)]
= [cosec² (360° – \(\overline{C+D}\))]²
= [- cosec (C + D)]²
= cosec² (C + D)

(v) since A + B + C + D = 2π
⇒ B + C + D = 2π – A
⇒ sin (B + C + D) = sin (2π – A)
= – sin A
⇒ sin A + sin (B + C + D) = 0

Question 16.
If A, B, C, D are angles of a cyclic quadrilateral, prove that
(i) cot A + cot B + cot C + cot D = 0
(ii) sin A + sin B = sin C + sin D.
Solution:
Since A, B, C and D are the angles of cyclic quadrilateral.
∴ A + C = B + D = π ………………..(1)

(i) L.H.S. = cot A + cot B + cot C + cot D
= cot A + cot B + cot (n – A) + cot (n – B) [using (1)]
= cot A + cot B – cot A – cot B = 0
= R.H.S.

(ii) L.H.S. = sin A + sin B
= sin (π – C) + sin (π – D) [using (1)]
= sin C + sin D
= R.H.S.
[∵ sin (π – θ) = sin θ]

Question 17.
Find tan 15° and hence show that tan 15° + cot 15° = 4.
Solution:
tan 150° = tan (45° – 30°)

Question 18.
Evaluate:
(i) cos 195°
(ii) sin \(\frac{11 \pi}{12}\)
Solution:
(i) cos 195° = cos (180° + 15°)
= – cos 15°
[∵ cos (180° – θ) = – cos θ]
= – cos (45° – 30°)
= – [cos 45° cos 30° + sin 45° sin 30°]
[∵ cos (A – B) = cos A cos B + sin A sin B]
= – \(\left[\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}\right]\)
= – \(\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)\)

(ii) sin \(\frac{11 \pi}{12}\)
= sin (π \(\frac{\pi}{12}\))
= sin \(\frac{\pi}{12}\)
=sin 15°
= sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
[∵ sin (A – B) = sin A cos B – cos A sin B]

Question 19.
Evaluate the following:
(i) cos 105° + sin 105°
(ii) cos 15° – sin 15°
(iii) cot 105° – tan 105°.
Solution:
(i) cos 105° + sin 105°
= cos (60° + 45°) + sin (60° + 45°)
= [cos 64° cos 45° – sin 60° sin 45°] + [sin 60° cos 45° + cos 60° sin 45°]
= \(\frac{1}{2} \times \frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2} \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{1}{\sqrt{2}}\)
= \(\frac{2}{2 \sqrt{2}}=\frac{1}{\sqrt{2}}\)

(ii) cos 15° = cos (45° – 30°)
= cos 45° cos 30° + sin 45° sin 30°
[∵ cos (A – B) = cos A cos B + sin A sin B]
= \(\frac{1}{\sqrt{2}} \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \frac{1}{2}\)
= \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)
sin 15° = sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
[∵ sin (A – B) = sin A cos B – cos A sin B]
= \(\frac{1}{\sqrt{2}} \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \frac{1}{2}\)
= \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)
∴ cos 15° – sin 15° = \(\frac{\sqrt{3}+1}{2 \sqrt{2}}-\frac{\sqrt{3}-1}{2 \sqrt{2}}\)
= \(\frac{\sqrt{3}+1-\sqrt{3}+1}{2 \sqrt{2}}\)
= \(\frac{2}{2 \sqrt{2}}=\frac{1}{\sqrt{2}}\)

(iii) cot 105° – tan 105° = cot (90° + 15°) – tan 105°
= – tan 15° – tan 105°
= – tan (45° – 30°) – tan (60° + 45°)

Question 20.
Show that:
(i) sin 38° cos 22° + cos 38° sin 22° = \(\frac{\sqrt{3}}{2}\)
(ii) sin 70° cos 10° – cos 70° sin 10° = \(\frac{\sqrt{3}}{2}\)
(iii) cos 130° cos 40° + sin 130° sin 40° = 0
(iv) \(\sin \frac{7 \pi}{12} \cos \frac{\pi}{4}-\cos \frac{7 \pi}{12} \sin \frac{\pi}{4}=\frac{\sqrt{3}}{2}\)
(v) \(\frac{\tan 69^{\circ}+\tan 66^{\circ}}{1-\tan 69^{\circ} \tan 66^{\circ}}\) = – 1
Solution:
(i) sin 38° cos 22° + cos 38° sin 22°
= sin (38° + 22°)
= sin 60°
= \(\frac{\sqrt{3}}{2}\)
[∵ sin (A – B) = sin A co sB – cos A sin B]

(ii) sin 70° cos 10° – cos 70° sin 10° = sin (70° – 10°)
= sin 60°
= \(\frac{\sqrt{3}}{2}\)
[∵ sin (A – B) = sin A cos B + cos A sin B

(iii) cos 130° cos40° + sin 130° sin 40°
= cos (130° – 40°)
= cos 90° = 0
[∵ cos (A – B) = cos A cos B + sin A sin B]

(iv) \(\sin \frac{7 \pi}{12} \cos \frac{\pi}{4}-\cos \frac{7 \pi}{12} \sin \frac{\pi}{4}\)
= sin \(\left(\frac{7 \pi}{12}-\frac{\pi}{4}\right)\)
= sin \(\left(\frac{4 \pi}{12}\right)\)
= sin \(\frac{\pi}{3}\)
= \(\frac{\sqrt{3}}{2}\)
[∵ sin A cos B – cos A sin B = sin (A – B)]

(v) \(\frac{\tan 69^{\circ}+\tan 66^{\circ}}{1-\tan 69^{\circ} \tan 66^{\circ}}\) = tan (69° + 66°)
= tan 135°
= tan (90° + 45°)
= – cot 45° = – 1
[∵ tan (A + B) = \(\frac{\tan A+\tan B}{1-\tan A \tan B}\)]

Question 21.
Prove that :
(i) sin (x – y) cos x – cos (x – y) sin x = – sin y
(ii) √2 cos (\(\frac{\pi}{4}\) + x) = cos x – sin x.
Solution:
(i) sin (x – y) cos x – cos (x – y) sin x = sin (x – y – x)
= sin (- y)
= – sin y
[∵ sin A cos B – cos A sin B = sin (A – B)]

(ii) L.H.S. = √2 cos (\(\frac{\pi}{4}\) + x)
= √2 [cos \(\frac{\pi}{4}\) cos x – sin \(\frac{\pi}{4}\) sin x]
[∵ cos (A + B) = cos A cos B – sin A sin B]
= √2 [\(\frac{1}{\sqrt{2}}\) cos x – \(\frac{1}{\sqrt{2}}\) sin x]
= cos x – sin x
= R.H.S.

Question 22.
(i) \(\cos ^2\left(\frac{\pi}{4}+x\right)-\sin ^2\left(\frac{\pi}{4}-x\right)\)
(ii) \(\sin ^2\left(\frac{\pi}{8}+\frac{x}{2}\right)-\sin ^2\left(\frac{\pi}{8}-\frac{x}{2}\right)\)
Soluion:
(i) \(\cos ^2\left(\frac{\pi}{4}+x\right)-\sin ^2\left(\frac{\pi}{4}-x\right)\)
= cos (\(\frac{\pi}{4}\) + x + \(\frac{\pi}{2}\) – x) cos (\(\frac{\pi}{2}\) + x – \(\frac{\pi}{2}\) + x)
[∵ cos² A – sin² B = cos (A + B) – cos (A – B)]
= cos \(\frac{\pi}{2}\) cos 2x = 0

(ii) \(\sin ^2\left(\frac{\pi}{8}+\frac{x}{2}\right)-\sin ^2\left(\frac{\pi}{8}-\frac{x}{2}\right)\)
= \(\sin \left(\frac{\pi}{8}+\frac{x}{2}+\frac{\pi}{8}-\frac{x}{2}\right) \sin \left(\frac{\pi}{8}+\frac{x}{2}-\frac{\pi}{8}+\frac{x}{2}\right)\)
[∵ sin² A – sin² B = sin (A + B) – sin (A – B)]
= sin \(\frac{\pi}{4}\) sin x
= \(\frac{1}{\sqrt{2}}\) sin x

Question 23.
(i) If sin A = \(\frac{1}{\sqrt{5}}\) and cos B = \(\frac{3}{\sqrt{10}}\), where A, B are positive acute angles, prove that A + B = 45°.
(ii) If cos A = \(\frac{1}{7}\) and cos B = \(\frac{13}{14}\) and A, B lie in first quadrant, prove that A – B = 60°.
Solution:
(i) Given sin A = \(\frac{1}{\sqrt{5}}\)

(ii) Given cos A = \(\frac{1}{7}\)
and cos B = \(\frac{13}{14}\)

Question 24.
(i) If α, β lie in first quadrant and sin α = \(\frac{8}{17}\), tan β = \(\frac{5}{12}\), find the values of sin (α – β), cos (α – β) and tan (α – β).
(ii) If cos x = \(\frac{4}{5}\), cos y = \(\frac{12}{13}\), \(\frac{3 \pi}{2}\) < x < 2π and \(\frac{3 \pi}{2}\) < y < 2π, find the values of cos (x +y) and sin (x – y).
Solution:
Given sin α = \(\frac{8}{17}\)
and tan β = \(\frac{5}{12}\)
∴ cos α = \(\sqrt{1-\sin ^2 \alpha}\)
= \(\sqrt{1-\frac{64}{289}}\)
= \(\sqrt{\frac{225}{289}}=\frac{15}{17}\)
[∵ α lies in first quadrant]
and sec² β = 1 + tan² β
= 1 + \(\left(\frac{5}{12}\right)^2\)
= 1 + \(\frac{25}{144}\)
= \(\frac{169}{144}\)
∴ sec β = ± \(\frac{13}{12}\)
since β lies in first quadrant
∴ sec β > 0
Thus, sec β = \(\frac{13}{12}\)

(ii)

Question 25.
(i) A positive acute angle is divided into two parts whose tangents are and Show that the angle is
(ii) Prove that tan 22° + tan 23° + tan 22° tan 23° = 1.
(iii) If α + β = \(\frac{\pi}{4}\), prove that (1 + tan α) (1 + tan β) = 2.
Solution:
(i) Let θ = α + β,
where tan α = \(\frac{1}{2}\)
and tan β = \(\frac{1}{3}\)
Now tan θ = tan (α + β)
= \(\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\)
= \(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}=\frac{\frac{5}{6}}{\frac{5}{6}}\)
⇒ tan θ = 1
⇒ B = \(\frac{\pi}{4}\)
Hence the required positive acute angle be \(\frac{\pi}{4}\).

(ii) Now tan 45° = (tan 22° + tan 23°)
= \(\frac{\tan 22^{\circ}+\tan 23^{\circ}}{1-\tan 22^{\circ} \tan 23^{\circ}}\)
⇒ 1 = \(\frac{\tan 22^{\circ}+\tan 23^{\circ}}{1-\tan 22^{\circ} \tan 23^{\circ}}\)
⇒ 1 – tan 22° tan 23° = tan 22° + tan 23°
⇒ 1 = tan 22° + tan 23° + tan 22° tan 23°

(iii) Given α + β = \(\frac{\pi}{4}\)
⇒ β = \(\frac{\pi}{4}\) – α
L.H.S. = (1 + tan α) (1 + tan β)

Question 26.
Prove that :
(i) tan 70° = tan 20° + 2 tan 50°
(ii) tan 56° = \(\frac{\tan 20^{\circ}+\tan 50^{\circ}}{1-\tan 20^{\circ} \tan 50^{\circ}}\)
Solution:
(i) tan 70° = tan (20° + 50°)
= \(\frac{\tan 20^{\circ}+\tan 50^{\circ}}{1-\tan 20^{\circ} \tan 50^{\circ}}\)
⇒ tan 70° [1 – tan 20° tan 50°]
= tan 20° + tan 50°
⇒ tan 70° – tan 20° tan 50° tan (90° — 70°)
= tan 20° + tan 50°
⇒ tan 70° – tan 20° tan 50° cot 20°
= tan 20° + tan 50°
⇒ tan 70° = tan 20° + 2 tan 50°

(ii) tan 56° = tan (45° + 11°)

Question 27.
Prove that :
(i) sin (x + y) sin (x – y) + sin (y + z) sin (y – z) + sin (z + x) sin (- x) = 0
(ii) 1 + tan x tan \(\frac{x}{2}\) = sec x = tan x cot \(\frac{x}{2}\) – 1.
Solution:
(i) L.H.S. = sin (x + y) sin (x – y) + sin (y + z) sin (y – z) + sin (z + x) sin (z – x)
= sin²x – sin² y + sin² y – sin² z + sin² z – sin² x
= 0
= R.HS.
[∵ sin (A + B) = sin (A – B) = sin² A – sin² B]

(ii) L.H.S. = 1 + tan x tan \(\frac{x}{2}\)

Question 28.
Prove that : tan 13x = tan 4x + tan 9x + tan 4x tan 9x tan 13x.
Solution:
Now, tan 13x = tan (4x + 9x)
= \(\frac{\tan 4 x+\tan 9 x}{1-\tan 4 x \tan 9 x}\)
⇒ tan 13x [1 – tan 4x tan 9x] = tan 4x + tan 9x
⇒ tan 13x = tan 4x + tan 9x + tan 4x tan 9x tan 13x

Question 29.
Prove that: cos 2x cos 2y + sin² (x – y) – sin² (x + y) = cos (2x + 2y).
Solution:
LH.S. = cos 2x cos 2y + sin² (x – y) – sin² (x + y)
= cos 2x cos 2y + sin (x – y + x + y) sin (x – y – x – y)
[∵ sin² A – sin² B = sin (A + B) sin (A – B)]
= cos 2x cos 2y + sin 2x sin (- 2y) .
= cos 2x cos 2y – sin 2x sin 2y
[∵ sin(- θ) = – sin θ]
= cos (2x + 2y)
[∵ cos A cos B – sin A sin B = cos (A + B)]
= R.H.S.

Question 30.
(i) cot x cot y = 2, show that \(\frac{\cos (x+y)}{\cos (x-y)}=\frac{1}{3}\).
(ii) If tan A = m tan B, prove that \(\frac{\sin (A-B)}{\sin (A+B)}=\frac{m-1}{m+1}\)
(ill) If α + β = γ and \(\frac{\tan \alpha}{\tan \beta}=\frac{x}{y}\), then prove that sin (α – β) = \(\frac{x-y}{x+y}\) sin γ.
Solution:
(i) Given cot x coty = 2
⇒ \(\frac{\cos x \cos y}{\sin x \sin y}=\frac{2}{1}\)
Applying componendo and dividendo ; we have
\(\frac{\cos x \cos y-\sin x \sin y}{\cos x \cos y+\sin x \sin y}=\frac{2-1}{2+1}\)
⇒ \(\frac{\cos (x+y)}{\cos (x-y)}=\frac{1}{3}\)

(ii) Given tan A = m tan B
⇒ \(\frac{\tan \mathrm{A}}{\tan \mathrm{B}}=\frac{m}{1}\)
⇒ \(\frac{\sin \mathrm{A} \cos \mathrm{B}}{\cos \mathrm{A} \sin \mathrm{B}}=\frac{m}{1}\)
Applying componendo and dividendo ; we have
⇒ \(\frac{\sin \mathrm{A} \cos \mathrm{B}-\cos \mathrm{A} \sin \mathrm{B}}{\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B}}=\frac{m-1}{m+1}\)
⇒ \(\frac{\sin (\mathrm{A}-\mathrm{B})}{\sin (\mathrm{A}+\mathrm{B})}=\frac{m-1}{m+1}\) ………………(1)

(iii) Given α + β = γ ……………….(1)

Question 31.
If 3 tan (θ – 15°) = tan (θ + 15°), 0 < θ < 90°, then prove that θ = 45°.
Solution:
Given, 3 tan (θ – 15°) = tan (θ + 15°)

Question 32.
Find the range of the function 5 sin x – 12 cos x + 7.
Solution:
Let f (x) = 5 sin x – 12 cos x + 7
putting 5 = r cos α
and 12 = rsin α …………….(2)
On squaring (1) and (2) ; we have
r = \(\sqrt{25+144}\)
= \(\sqrt{169}\) = 13
On dividing (2) by (1) ; we have
tan α = \(\frac{12}{5}\)
∴ f (x) = r sin x cos α – r sin α cos x + 7
= 13 sin (x – α) + 7
since – 13 ≤ sin (x – α) ≤ 1 ∀ x ∈ R
⇒ – 13 ≤ 13 sin (x – α) ≤ 13
⇒ – 13 + 7 ≤ 13 sin (x- α) + 7 ≤ 13 + 7
⇒ – 6 ≤ f (x) ≤ 20
This range of f (x) = R_f = [- 6, 20]