Trigonometry – Trigonometric Identities – JEE Main 24 Jan 2026 Shift 1

Solution:

Let $N$ be the numerator and $D$ be the denominator.

**Solving Numerator ($N$):**
$$N = \sqrt{3} cosec 20^{\circ} – \sec 20^{\circ} = \frac{\sqrt{3}}{\sin 20^{\circ}} – \frac{1}{\cos 20^{\circ}}$$
$$N = \frac{\sqrt{3}\cos 20^{\circ} – \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$$
Multiply numerator and denominator by 2:
$$N = \frac{2\left(\frac{\sqrt{3}}{2}\cos 20^{\circ} – \frac{1}{2}\sin 20^{\circ}\right)}{\frac{1}{2}(2\sin 20^{\circ} \cos 20^{\circ})}$$
$$N = \frac{4(\sin 60^{\circ}\cos 20^{\circ} – \cos 60^{\circ}\sin 20^{\circ})}{\sin 40^{\circ}}$$
$$N = \frac{4\sin(60^{\circ}-20^{\circ})}{\sin 40^{\circ}} = \frac{4\sin 40^{\circ}}{\sin 40^{\circ}} = 4$$

**Solving Denominator ($D$):**
$$D = \cos 20^{\circ}\cos 40^{\circ}\cos 60^{\circ}\cos 80^{\circ}$$
Since $\cos 60^{\circ} = \frac{1}{2}$:
$$D = \frac{1}{2} (\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ})$$
Using the identity $\cos \theta \cos 2\theta \cos 2^2 \theta \dots \cos 2^{n-1}\theta = \frac{\sin(2^n \theta)}{2^n \sin \theta}$ with $n=3, \theta=20^{\circ}$:
$$D = \frac{1}{2} \left[ \frac{\sin(2^3 \times 20^{\circ})}{2^3 \sin 20^{\circ}} \right] = \frac{1}{2} \left[ \frac{\sin 160^{\circ}}{8 \sin 20^{\circ}} \right]$$
Since $\sin 160^{\circ} = \sin(180^{\circ}-20^{\circ}) = \sin 20^{\circ}$:
$$D = \frac{1}{2} \cdot \frac{\sin 20^{\circ}}{8 \sin 20^{\circ}} = \frac{1}{16}$$

**Final Calculation:**
$$\text{Expression} = \frac{N}{D} = \frac{4}{1/16} = 4 \times 16 = 64$$

Ans. (3)