Trigonometry – Trigonometric Ratios and Identities – JEE Main 21 Jan 2026 Shift 1

Solution:

Expression: $\frac{1}{\sin 10^{\circ}} – \frac{\sqrt{3}}{\cos 10^{\circ}}$

Take LCM:
$$= \frac{\cos 10^{\circ} – \sqrt{3} \sin 10^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$$

Multiply and divide the numerator by 2:
$$= \frac{2 \left( \frac{1}{2} \cos 10^{\circ} – \frac{\sqrt{3}}{2} \sin 10^{\circ} \right)}{\sin 10^{\circ} \cos 10^{\circ}}$$
$$= \frac{2 ( \sin 30^{\circ} \cos 10^{\circ} – \cos 30^{\circ} \sin 10^{\circ} )}{\sin 10^{\circ} \cos 10^{\circ}}$$
Using $\sin(A-B) = \sin A \cos B – \cos A \sin B$:
$$= \frac{2 \sin(30^{\circ} – 10^{\circ})}{\sin 10^{\circ} \cos 10^{\circ}}$$
$$= \frac{2 \sin 20^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$$

Multiply numerator and denominator by 2 to form $\sin 2\theta$ in the denominator:
$$= \frac{2(2 \sin 20^{\circ})}{2 \sin 10^{\circ} \cos 10^{\circ}}$$
$$= \frac{4 \sin 20^{\circ}}{\sin 20^{\circ}}$$
$$= 4$$

Ans. (1)