Class 10, Maths, Chapter 8, Exercise 8.2, Solutions
Q.1. Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan2 45° + cos2 30° – sin2 60°
(iii) $\frac{\cos \,\,{{45}^{0}}}{\sec \,\,{{30}^{0}}+\cos ec\,\,{{30}^{0}}}$
(iv) $\frac{\sin \,{{30}^{0}}+\tan \,{{45}^{0}}-\cos ec\,{{60}^{0}}}{\sec \,{{30}^{0}}+\cos \,{{60}^{0}}+\cot \,{{45}^{0}}}$
(v) $\frac{5\,{{\cos }^{2}}\,\,{{60}^{0}}+4\,\,{{\sec }^{2}}\,{{30}^{0}}-{{\tan }^{2}}\,\,{{45}^{0}}}{{{\sin }^{2}}\,{{30}^{0}}+{{\cos }^{2}}\,{{30}^{0}}}$
Ans:
Q.2. Choose the correct option and justify your choice:
(i) $\frac{2\,\,\tan \,\,{{30}^{0}}}{1+{{\tan }^{2}}\,\,{{30}^{0}}}=$
(A) (B) cos 60° (C) tan 60° (D) sin 30°
(ii) $\frac{1-{{\tan }^{2}}\,\,{{45}^{0}}}{1+{{\tan }^{2}}\,\,{{45}^{0}}}=$
(A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) sin 2A = 2 sin A is true when A=
(A) 00 (B) 300 (C) 450 (d) 600
(iv) $\frac{2\,\,\tan \,\,{{30}^{0}}}{1-{{\tan }^{2}}\,\,{{30}^{0}}}=$
- cos 60° (B) sin 60° (C) tan 60° (D) sin 30°
Ans: (A)
(ii) Correct option: (D)
Justification,
$\frac{1-{{\tan }^{2}}\,\,{{45}^{0}}}{1+{{\tan }^{2}}\,\,{{45}^{0}}}=\frac{1-1}{1+1}=\frac{0}{2}=0$
(iii) Correct option: (A)
Justification,
when A=00,
sin 2A = sin (2 × 00) = 0
And, 2 sin A = 2 sin 00 = (2 × 0) = 0
⇒ sin 2A = 2 sin A, when A=0
(iv) correct answer (C)
Q.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.
Ans:
Q.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 θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Ans: (i) False
(ii) True
(iii) False
(iv) False
Justification,
⇒ sin 900 = cos 900
⇒ 1≠0
it is only true for $\theta -{{45}^{0}}\left( \sin \,\,{{45}^{0}}=\frac{1}{\sqrt{2}}=\cos \,\,{{45}^{0}} \right)$
(v) True
Justification,
⇒ cot 00 = 0 and cot 00 = $\frac{1}{\tan \,\,{{0}^{0}}}=\frac{1}{0}$ i.e., Not defined.