# Bihar Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2

Bihar Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2 Textbook Questions and Answers.

## BSEB Bihar Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2

Question 1.
Evaluate :
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan2 45° + cos2 30° – sin2 60°

Solution:
(i) sin 60° cos 30° + sin 30° cos 60°
= $$\frac{\sqrt{3}}{2}$$ x $$\frac{\sqrt{3}}{2}$$ + $$\frac { 1 }{ 2 }$$ x $$\frac { 1 }{ 2 }$$
= $$\frac { 3 }{ 4 }$$ + $$\frac { 1 }{ 4 }$$ = $$\frac { 3+1 }{ 4 }$$ = $$\frac { 4 }{ 4 }$$ = 1.

(ii) 2 tan² 45° + cos² 30° – sin² 60°
= 2(1)² + $$\left(\frac{\sqrt{3}}{2}\right)^{2}$$ – $$\left(\frac{\sqrt{3}}{2}\right)^{2}$$
= 2 + $$\frac { 3 }{ 4 }$$ – $$\frac { 3 }{ 4 }$$ = 2

Question 2.
Choose the correct option and justify:
(i) $$\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}$$
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°

(ii) $$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}$$
(A) tan 90°
(B) 1
(C) sin 45°
(D) O

(iii) sin 2A = 2 sin A is true when A =
(A) 0°
(B) 30°
(C) 45°
(D) 60°

(iv) $$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}$$
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) none of these
Solution:
(i) (A)

(ii) (D)
Because $$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}$$ = $$\frac { 1-1 }{ 1+1 }$$ = $$\frac { 0 }{ 2 }$$ = 0

(iii) (A)
Because when A – 0°, sin 2A = sin 0° = 0
and, 2 sin A = 2 sin 0°= 2 x 0 = 0
or sin 2 A = 2 sin A, when A = 0°

(iv) (C)

Question 3.
If tan (A + B) = $$\sqrt{3}$$ and tan (A – B) = $$\frac{1}{\sqrt{3}}$$; 0° < A + B ≤ 90°; A > B, find A and B.
Solution:
tan (A + B) = $$\sqrt{3}$$ means tan (A + B) = tan 60°
So, A + B = 60° … (1)
tan (A – B) = $$\frac{1}{\sqrt{3}}$$ means tan (A – B) = tan 30°
So, A – B = 30° … (2)
Solving (1) and (2), we get
A = 45° and B = 15°
Hence, A = 45° and B = 15°.

Question 4.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin 0 = cos 0 for all values of 0.
(v) cot A is not defined for A = 0°.
Solution:
(i) False. Because
When A = 60° and B = 30°. Then,
sin (A + B) = sin (60° + 30°) = sin 90° = 1
and, sin A + sin B = sin 60° + sin 30°
= $$\frac{\sqrt{3}}{2}$$ + $$\frac { 1 }{ 2 }$$ = $$\frac{\sqrt{3}+1}{2}$$
So, sin (A + B) ≠ sin A + sin B

(ii) True. Because, it is clear from the table below :

that the value of sin θ increases as θ increases.

(iii) False. Because it is clear from the table below :

that the value of cos θ decreases as θ increases.

(iv) False. Because it is only true for θ = 45°.
(sin 45° = $$\frac{1}{\sqrt{2}}$$ = cos 45°)

(v) True. Because tan 0° = 0 and
cot 0° = $$\frac{1}{\tan 0^{\circ}}$$ = $$\frac { 1 }{ 0 }$$, i.e., not defined.