NCERT Class 10 Maths – Exercise 8.2 Solutions

NCERT Class 10 Maths

Chapter 8 – Introduction to Trigonometry | Exercise 8.2

(Rationalized Syllabus 2025-26)

💡 Trigonometric Ratios Table

Angle0°30°45°60°90°
sin A01/21/√2√3/21
cos A1√3/21/√21/20
tan A01/√31√3Not Defined
Q1

Evaluate the following:

(i) $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$
Substitute values:
$$= \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$$
$$= \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1$$
✔️ 1
(ii) $2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ$
Substitute values:
$$= 2(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 – \left(\frac{\sqrt{3}}{2}\right)^2$$
$$= 2(1) + \frac{3}{4} – \frac{3}{4} = 2$$
✔️ 2
(iii) $\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ}$
Substitute values:
$$= \frac{1/\sqrt{2}}{2/\sqrt{3} + 2} = \frac{1/\sqrt{2}}{(2 + 2\sqrt{3})/\sqrt{3}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2(1 + \sqrt{3})}$$
$$= \frac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})}$$
Rationalize denominator:
$$= \frac{\sqrt{3}(\sqrt{3}-1)}{2\sqrt{2}(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{3 – \sqrt{3}}{2\sqrt{2}(3-1)} = \frac{3-\sqrt{3}}{4\sqrt{2}}$$
Multiply by $\sqrt{2}/\sqrt{2}$:
$$= \frac{(3-\sqrt{3})\sqrt{2}}{4\sqrt{2}\cdot\sqrt{2}} = \frac{3\sqrt{2} – \sqrt{6}}{8}$$
✔️ $\frac{3\sqrt{2} – \sqrt{6}}{8}$
(iv) $\frac{\sin 30^\circ + \tan 45^\circ – \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$
Substitute values:
$$= \frac{\frac{1}{2} + 1 – \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1} = \frac{\frac{3}{2} – \frac{2}{\sqrt{3}}}{\frac{3}{2} + \frac{2}{\sqrt{3}}}$$
$$= \frac{\frac{3\sqrt{3}-4}{2\sqrt{3}}}{\frac{3\sqrt{3}+4}{2\sqrt{3}}} = \frac{3\sqrt{3}-4}{3\sqrt{3}+4}$$
Rationalize:
$$= \frac{(3\sqrt{3}-4)^2}{(3\sqrt{3})^2 – 4^2} = \frac{27 + 16 – 24\sqrt{3}}{27 – 16} = \frac{43 – 24\sqrt{3}}{11}$$
✔️ $\frac{43 – 24\sqrt{3}}{11}$
(v) $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ – \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$
Note: Denominator $\sin^2 30^\circ + \cos^2 30^\circ = 1$.
$$= 5(\frac{1}{2})^2 + 4(\frac{2}{\sqrt{3}})^2 – (1)^2$$
$$= 5(\frac{1}{4}) + 4(\frac{4}{3}) – 1 = \frac{5}{4} + \frac{16}{3} – 1$$
$$= \frac{15 + 64 – 12}{12} = \frac{67}{12}$$
✔️ 67/12
Q2

Choose the correct option and justify your choice:

(i) $\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} =$
Substitute $\tan 30^\circ = 1/\sqrt{3}$:
$$= \frac{2(1/\sqrt{3})}{1 + (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{1 + 1/3} = \frac{2/\sqrt{3}}{4/3}$$
$$= \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{\sqrt{3}}{2}$$
Since $\sin 60^\circ = \sqrt{3}/2$:
✔️ (A) $\sin 60^\circ$
(ii) $\frac{1 – \tan^2 45^\circ}{1 + \tan^2 45^\circ} =$
Substitute $\tan 45^\circ = 1$:
$$= \frac{1 – 1^2}{1 + 1^2} = \frac{0}{2} = 0$$
✔️ (D) 0
(iii) $\sin 2A = 2 \sin A$ is true when A =
Test for $A = 0^\circ$:
$$\text{LHS} = \sin(2 \times 0) = \sin 0 = 0$$
$$\text{RHS} = 2 \sin 0 = 2(0) = 0$$
LHS = RHS.
✔️ (A) 0°
(iv) $\frac{2 \tan 30^\circ}{1 – \tan^2 30^\circ} =$
Substitute $\tan 30^\circ = 1/\sqrt{3}$:
$$= \frac{2(1/\sqrt{3})}{1 – (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{1 – 1/3} = \frac{2/\sqrt{3}}{2/3}$$
$$= \frac{2}{\sqrt{3}} \times \frac{3}{2} = \sqrt{3}$$
Since $\tan 60^\circ = \sqrt{3}$:
✔️ (C) $\tan 60^\circ$
Q3

If $\tan(A + B) = \sqrt{3}$ and $\tan(A – B) = 1/\sqrt{3}$; $0^\circ < A + B \le 90^\circ$; $A > B$, find A and B.

Given $\tan(A + B) = \sqrt{3}$. Since $\tan 60^\circ = \sqrt{3}$:
$$A + B = 60^\circ \quad \text{…(1)}$$
Given $\tan(A – B) = 1/\sqrt{3}$. Since $\tan 30^\circ = 1/\sqrt{3}$:
$$A – B = 30^\circ \quad \text{…(2)}$$
Adding equations (1) and (2):
$$2A = 90^\circ \Rightarrow A = 45^\circ$$
Substitute $A = 45^\circ$ in (1):
$$45^\circ + B = 60^\circ \Rightarrow B = 15^\circ$$
✔️ A = 45°, B = 15°
Q4

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

(i) $\sin(A + B) = \sin A + \sin B$.
Let $A=30^\circ, B=60^\circ$.
LHS = $\sin(90^\circ) = 1$. RHS = $\sin 30^\circ + \sin 60^\circ = 0.5 + 0.866 \neq 1$.
False
(ii) The value of $\sin \theta$ increases as $\theta$ increases.
Values: $\sin 0^\circ=0, \sin 30^\circ=0.5, \sin 45^\circ \approx 0.7, \sin 60^\circ \approx 0.87, \sin 90^\circ=1$.
True
(iii) The value of $\cos \theta$ increases as $\theta$ increases.
Values: $\cos 0^\circ=1, \cos 90^\circ=0$. It decreases.
False
(iv) $\sin \theta = \cos \theta$ for all values of $\theta$.
Only true for $\theta = 45^\circ$. For $\theta=30^\circ, \sin 30 \neq \cos 30$.
False
(v) $\cot A$ is not defined for $A = 0^\circ$.
$\cot 0^\circ = \frac{\cos 0^\circ}{\sin 0^\circ} = \frac{1}{0}$ (Undefined).
True
🎉 Exercise 8.2 Completed
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