Class 10 - Chapter 8: Introduction to Trigonometry

Section A (1 Mark each)

Q1. If $\sin A = \frac{3}{5}$, find $\tan A$.

(a) $\frac{3}{4}$ (b) $\frac{4}{3}$ (c) $\frac{5}{3}$ (d) $\frac{4}{5}$

Q2. Evaluate: $\cos 48^\circ - \sin 42^\circ$.

(a) 1 (b) 0 (c) -1 (d) 2

Q3. Given $15 \cot A = 8$, find $\sec A$.

(a) $\frac{17}{8}$ (b) $\frac{8}{17}$ (c) $\frac{15}{8}$ (d) $\frac{8}{15}$

Q4. Assertion (A): The value of $\sin A$ increases as $A$ increases for $0^\circ < A < 90^\circ$.
Reason (R): The value of $\cos A$ increases as $A$ increases for $0^\circ < A < 90^\circ$.

(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.

Section B (2 Marks each)

Q5. If $\sec \theta + \tan \theta = p$, then prove that $\sin \theta = \frac{p^2 - 1}{p^2 + 1}$.

Solve on white paper.

Q6. Prove that $\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}$.

Solve on white paper.

Q7. If $\tan(A+B) = \sqrt{3}$ and $\tan(A-B) = \frac{1}{\sqrt{3}}$; $0^\circ < A+B \le 90^\circ$; $A > B$, find A and B.

Solve on white paper.

Section C (3 Marks each)

Q8. Prove that: $\frac{\cos A}{1+\sin A} + \frac{1+\sin A}{\cos A} = 2 \sec A$.

Solve on white paper.

Q9. If $\sin \theta + \cos \theta = \sqrt{2}$, prove that $\tan \theta + \cot \theta = 2$.

Solve on white paper.

Section D (5 Marks)

Q10. If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n$, prove that $m^2 - n^2 = 4\sqrt{mn}$.

Solve on white paper.

Section E (4 Marks)

Q11. Case Study:

In a right-angled triangle ABC, right-angled at B, AB = 24 cm, BC = 7 cm.

(i) Find $\sin A$ and $\cos A$.

(ii) Find $\sin C$ and $\cos C$.

(iii) Find $\tan A$.

Solve on white paper.