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

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 = $$\frac{\pi}{4}$$