Introduction to Trigonometry: Class 10 Important Questions with Answers

HOTS (higher-order) practice

H1. If $\sec \theta +\tan \theta =p$, find $\sec \theta -\tan \theta$ and hence $\sin \theta$ in terms of p.

H2. If $\tan \theta +\cot \theta =2$, find ${\tan }^2\theta +{\cot }^2\theta$.

Case-based practice

In 2023–2025 the case studies that used trigonometry were all heights-and-distances (Chapter 9). These two practise this chapter's ideas in the board's case format.

20 MCQs (1 mark each)

1. If $\sin A=\frac{3}{5}$, then $\cos A$ is
   (a) $\frac{4}{5}$ (b) $\frac{3}{4}$ (c) $\frac{5}{4}$ (d) $\frac{5}{3}$
2. $\sin 30°+\cos 60°$ equals
   (a) 0 (b) 1 (c) $\frac{1}{2}$ (d) $\sqrt{3}$
3. $\tan 45°\cdot \cot 45°$ is
   (a) 0 (b) 2 (c) 1 (d) not defined
4. ${\sec }^{2}A-{\tan }^{2}A$ is
   (a) 0 (b) 1 (c) −1 (d) 2
5. If $\tan \theta =1$, then θ is
   (a) 30° (b) 60° (c) 45° (d) 90°
6. $\frac{\sin 60°}{\cos 60°}$ equals
   (a) $\frac{1}{\sqrt{3}}$ (b) $\sqrt{3}$ (c) 1 (d) $\frac{\sqrt{3}}{2}$
7. Which is not defined?
   (a) tan 0° (b) cot 90° (c) sec 90° (d) cos 90°
8. $(1-{\sin }^{2}A){\sec }^{2}A$ is
   (a) 0 (b) 1 (c) ${\tan }^{2}A$ (d) ${\sin }^{2}A$
9. If $\cos A=\frac{1}{2}$, then $\tan A$ is
   (a) $\sqrt{3}$ (b) $\frac{1}{\sqrt{3}}$ (c) 1 (d) 2
10. ${\sin }^{2}45°+{\cos }^{2}45°$ is
    (a) $\frac{1}{2}$ (b) 2 (c) 1 (d) 0
11. If $5\tan \theta =4$, then $\frac{5\sin \theta -3\cos \theta }{5\sin \theta +2\cos \theta }$ is
    (a) $\frac{1}{6}$ (b) $\frac{1}{4}$ (c) $\frac{1}{2}$ (d) $\frac{5}{6}$
12. $\cos 0°+\sin 90°+\sqrt{2}\sin 45°$ is
    (a) 2 (b) 3 (c) $2+\sqrt{2}$ (d) 1
13. $\frac{\cos A}{\sin A}$ is
    (a) tan A (b) sec A (c) cot A (d) cosec A
14. $(\text{cosec}A-\cot A)(\text{cosec}A+\cot A)$ is
    (a) 0 (b) 1 (c) 2 (d) ${\text{cosec}}^{2}A$
15. If A + B = 90°, $\sin A=\frac{1}{2}$, then $\cos B$ is
    (a) $\frac{1}{2}$ (b) $\frac{\sqrt{3}}{2}$ (c) 1 (d) 0
16. $2\sin 30°\cos 30°$ equals
    (a) sin 60° (b) cos 60° (c) tan 60° (d) 1
17. $\sqrt{1+{\tan }^{2}A}$ (A acute) is
    (a) sec A (b) cos A (c) cosec A (d) tan A
18. The maximum value of $\sin \theta$ for acute θ (0° ≤ θ ≤ 90°) is
    (a) $\frac{1}{2}$ (b) 2 (c) 1 (d) $\sqrt{3}$
19. ${\tan }^{2}60°-{\sec }^{2}30°$ is
    (a) $\frac{5}{3}$ (b) 1 (c) $\frac{4}{3}$ (d) 3
20. If $\sin \theta =\cos \theta$ (acute), then $2\tan \theta +{\cos }^2\theta$ is
    (a) $\frac{5}{2}$ (b) 2 (c) $\frac{3}{2}$ (d) 3

Answer key

1 (a) · 2 (b) · 3 (c) · 4 (b) · 5 (c) · 6 (b) · 7 (c) · 8 (b) · 9 (a) · 10 (c) · 11 (a) · 12 (b) · 13 (c) · 14 (b) · 15 (a) · 16 (a) · 17 (a) · 18 (c) · 19 (a) · 20 (a)

Hints: Q11 $\frac{5\tan -3}{5\tan +2}=\frac{1}{6}$. Q12 $1+1+1$. Q15 B = 60°, cos 60° = ½. Q16 $2\cdot \frac{1}{2}\cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$. Q19 $3-\frac{4}{3}$. Q20 θ = 45°: $2+\frac{1}{2}$.

Assertion–Reason (10)

(a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.

1. A: ${\sin }^{2}67°+{\cos }^{2}67°=1$. R: ${\sin }^2\theta +{\cos }^2\theta =1$ for every θ.
2. A: $\sin \theta =\frac{5}{4}$ for some acute θ. R: The hypotenuse is the longest side of a right triangle.
3. A: tan 90° is not defined. R: cos 90° = 0.
4. A: $\sec \theta =\frac{12}{5}$ is possible. R: sec θ ≥ 1 for acute θ.
5. A: $\sin (A+B)=\sin A+\sin B$. R: sin is an increasing function on 0° to 90°.
6. A: $\frac{1-{\tan }^{2}45°}{1+{\tan }^{2}45°}=0$. R: tan 45° = 1.
7. A: cos 30° > cos 60°. R: cos θ decreases as θ increases from 0° to 90°.
8. A: If $\tan A=\frac{4}{3}$, then $\sin A=\frac{4}{5}$. R: $\sin A=\frac{\tan A}{\sqrt{1+{\tan }^{2}A}}$.
9. A: $\sin 2A=2\sin A$ is not true for A = 30°. R: $\sin 60°=\frac{\sqrt{3}}{2}$ while $2\sin 30°=1$.
10. A: ${\sec }^2\theta -{\tan }^2\theta =1$. R: $1+{\cot }^2\theta ={\text{cosec}}^2\theta$.

Key: 1 (a) · 2 (d) · 3 (a) · 4 (a) · 5 (d) · 6 (a) · 7 (a) · 8 (a) · 9 (a) · 10 (b)

Q2: sin θ ≤ 1 because the opposite side is shorter than the hypotenuse, so A is false. Q8: $\frac{4/3}{5/3}=\frac{4}{5}$ ✓. Q10: both true, but R is a different identity ⇒ (b).

Very short answer (1 mark each)

1. Find ${\sin }^{2}60°+{\cos }^{2}30°$. Ans. $\frac{3}{4}+\frac{3}{4}=\frac{3}{2}$
2. If $\cot \theta =\frac{1}{\sqrt{3}}$, find θ. Ans. 60°
3. Write $\tan A$ in terms of $\sin A$. Ans. $\frac{\sin A}{\sqrt{1-{\sin }^{2}A}}$
4. Evaluate $\frac{\tan 30°}{\cot 60°}$. Ans. 1
5. If $\sin A=\cos A$, find A. Ans. 45°
6. Find $9{\sec }^{2}A-9{\tan }^{2}A$. Ans. 9
7. If $\sec \theta =2$, find θ. Ans. 60°
8. Find $\sin 45°\cos 45°$. Ans. $\frac{1}{2}$

Short answer practice (2 marks each)

1. If $\sin (A+B)=1$ and $\cos (A-B)=\frac{\sqrt{3}}{2}$, with $0°<A+B\le 90°$ and A > B, find A and B.

2. If $\tan \theta =\frac{a}{b}$, show that $\frac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta }=\frac{a^2-b^2}{a^2+b^2}$.

3. Evaluate $\frac{\sin 30°+\tan 45°-\text{cosec}60°}{\sec 30°+\cos 60°+\cot 45°}$.

Short answer practice (3 marks each)

4. Prove that $\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\sin A+\cos A$.

5. Prove that $(\sin A+\sec A{)}^2+(\cos A+\text{cosec}A{)}^2=(1+\sec A\text{cosec}A{)}^2$.

6. Prove that $\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}=\frac{1+\sin \theta }{\cos \theta }$.

7. If $\sin \theta +{\sin }^2\theta =1$, prove that ${\cos }^2\theta +{\cos }^4\theta =1$.

8. Prove that $\sec A(1-\sin A)(\sec A+\tan A)=1$. CBSE 2023 · 30/4/1