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.
State whether the following are true or false. Justify your answer.
(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.