JEE Main 2026 — 24 January, Morning Shift — Question 8

Let $E = \frac{\sqrt{3}\csc 20^\circ - \sec 20^\circ}{\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ}$. $\sqrt{3}\csc 20^\circ - \sec 20^\circ = \frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ} = \frac{\sqrt{3}\cos 20^\circ - \sin 20^\circ}{\sin 20^\circ \cos 20^\circ}$. Multiply numerator and denominator by 2: $= \frac{2(\sqrt{3}\cos 20^\circ - \sin 20^\circ)}{2\sin 20^\circ \cos 20^\circ} = \frac{2(\sqrt{3}\cos 20^\circ - \sin 20^\circ)}{\sin 40^\circ}$. Now, $\sqrt{3}\cos 20^\circ - \sin 20^\circ = 2\left(\frac{\sqrt{3}}{2}\cos 20^\circ - \frac{1}{2}\sin 20^\circ\right) = 2\sin(60^\circ - 20^\circ) = 2\sin 40^\circ$. Thus numerator becomes $\frac{2 \cdot 2\sin 40^\circ}{\sin 40^\circ} = 4$. Denominator: $\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{2} \cos 20^\circ \cos 40^\circ \cos 80^\circ$. Using identity $\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8}$,

we get denominator = $\frac{1}{2} \cdot \frac{1}{8} = \frac{1}{16}$. Therefore, $E = \frac{4}{1/16} = 64$.