JEE Main 2026 Jan 24 Shift 1, Mathematics Q8: Trigonometric Ratios And Identities

Simplify the numerator:

$\sqrt{3}\csc 20^{\circ }-\sec 20^{\circ }=\frac{\sqrt{3}\cos 20^{\circ }-\sin 20^{\circ }}{\sin 20^{\circ }\cos 20^{\circ }}$

$=\frac{2(\sin 60^{\circ }\cos 20^{\circ }-\cos 60^{\circ }\sin 20^{\circ })}{\sin 20^{\circ }\cos 20^{\circ }}=\frac{2\sin 40^{\circ }}{\frac{1}{2}\sin 40^{\circ }}=4$.

Simplify the denominator using $\cos \theta \cos (60^{\circ }-\theta )\cos (60^{\circ }+\theta )=\frac{1}{4}\cos 3\theta$ with $\theta =20^{\circ }$:

$\cos 20^{\circ }\cos 40^{\circ }\cos 80^{\circ }=\frac{1}{4}\cos 60^{\circ }=\frac{1}{8}$.

Multiplying by $\cos 60^{\circ }=\frac{1}{2}$ gives denominator $=\frac{1}{16}$.

Result: $\frac{4}{1/16}=64$.