NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.2#
Exercise 8.2 deals with trigonometric ratios of specific angles: 0°, 30°, 45°, 60°, and 90°. Students use these standard values to evaluate expressions.
Standard Trigonometric Values:
| Angle | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan | 0 | 1/√3 | 1 | √3 | ∞ |
| cot | ∞ | √3 | 1 | 1/√3 | 0 |
| sec | 1 | 2/√3 | √2 | 2 | ∞ |
| csc | ∞ | 2 | √2 | 2/√3 | 1 |
1. Evaluate the following: #
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan²45° + cos²30° – sin²60°
(iii) $\dfrac{\cos 45°}{\sec 30° + \csc 30°}$
Rationalizing: multiply by $(\sqrt{6}-\sqrt{2})$ over itself…
$$= \frac{\sqrt{3}(2\sqrt{6}-2\sqrt{2})}{(2\sqrt{2}+2\sqrt{6})(2\sqrt{6}-2\sqrt{2})} = \frac{2\sqrt{18}-2\sqrt{6}}{4(6-2)} = \frac{6\sqrt{2}-2\sqrt{6}}{16} = \frac{3\sqrt{2}-\sqrt{6}}{8}$$∴ $= \dfrac{3\sqrt{2}-\sqrt{6}}{8}$
(iv) $\dfrac{\sin 30° + \tan 45° - \csc 60°}{\sec 30° + \cos 60° + \cot 45°}$
Numerator:
$$\sin 30° + \tan 45° - \csc 60° = \frac{1}{2} + 1 - \frac{2}{\sqrt{3}} = \frac{3}{2} - \frac{2\sqrt{3}}{3} = \frac{9-4\sqrt{3}}{6}$$Denominator:
$$\sec 30° + \cos 60° + \cot 45° = \frac{2}{\sqrt{3}} + \frac{1}{2} + 1 = \frac{2\sqrt{3}}{3} + \frac{3}{2} = \frac{4\sqrt{3}+9}{6}$$$$= \frac{9-4\sqrt{3}}{9+4\sqrt{3}}$$Rationalize by multiplying by $(9-4\sqrt{3})$:
$$= \frac{(9-4\sqrt{3})^2}{(9)^2-(4\sqrt{3})^2} = \frac{81-72\sqrt{3}+48}{81-48} = \frac{129-72\sqrt{3}}{33} = \frac{43-24\sqrt{3}}{11}$$(v) $\dfrac{5\cos^2 60° + 4\sec^2 30° - \tan^2 45°}{\sin^2 30° + \cos^2 30°}$
(Since $\sin^2 30° + \cos^2 30° = 1$)
$$= \frac{\frac{5}{4} + \frac{16}{3} - 1}{1} = \frac{15 + 64 - 12}{12} = \frac{67}{12}$$2. Choose the correct option and justify your choice: #
(i) $\dfrac{2\tan 30°}{1+\tan^2 30°}$ = (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
Answer: (A) sin 60°
(ii) $\dfrac{1-\tan^2 45°}{1+\tan^2 45°}$ = (A) tan 90° (B) 1 (C) sin 45° (D) 0
Answer: (D) 0
(iii) sin 2A = 2 sin A is true when A = (A) 0° (B) 30° (C) 45° (D) 60°
$\sin 2A = 2\sin A$
$2\sin A\cos A = 2\sin A$
$\cos A = 1$ (if $\sin A \neq 0$)
$A = 0°$ satisfies this (but then $\sin A = 0$ too, so check: $\sin 0° = 0$ and $2\sin 0° = 0$: $0 = 0$ ✓).
Answer: (A) 0°
(iv) $\dfrac{2\tan 30°}{1-\tan^2 30°}$ = (A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°
Answer: (C) tan 60°
3. If tan (A + B) = √3 and tan (A – B) = $\dfrac{1}{\sqrt{3}}$; 0° < A + B ≤ 90°; A > B, find A and B. #
Answer
$\tan(A+B) = \sqrt{3} = \tan 60°$
$$\Rightarrow A + B = 60° \quad \cdots (1)$$$\tan(A-B) = \dfrac{1}{\sqrt{3}} = \tan 30°$
$$\Rightarrow A - B = 30° \quad \cdots (2)$$Adding (1) and (2): $2A = 90° \Rightarrow A = 45°$
From (1): $B = 60° - 45° = 15°$
∴ $A = 45°$ and $B = 15°$.
4. State whether the following are true or false. Justify your answer. #
(i) sin (A + B) = sin A + sin B
False.
Example: $A = B = 30°$
$\sin(30°+30°) = \sin 60° = \dfrac{\sqrt{3}}{2}$
$\sin 30° + \sin 30° = \dfrac{1}{2} + \dfrac{1}{2} = 1$
$\dfrac{\sqrt{3}}{2} \neq 1$
So the statement is false.
(ii) The value of sin θ increases as θ increases.
True.
As $\theta$ increases from 0° to 90°, $\sin\theta$ increases from 0 to 1.
(iii) The value of cos θ increases as θ increases.
False.
As $\theta$ increases from 0° to 90°, $\cos\theta$ decreases from 1 to 0.
(iv) sin θ = cos θ for all values of θ.
False.
$\sin\theta = \cos\theta$ only when $\theta = 45°$. For example, $\sin 0° = 0 \neq 1 = \cos 0°$.
(v) cot A is not defined for A = 0°.
True.
$\cot A = \dfrac{\cos A}{\sin A}$. At $A = 0°$: $\sin 0° = 0$, so $\cot 0°$ is undefined (division by zero).