The value of cosec 10 ° -√(3) sec 10 ° is equal to:

Mathematics · Trigonometry Ratios and Identities

JEE Main 2026 — 21 January, Morning Shift — Question 18

The value of $\operatorname{cosec} 10^{\circ}-\sqrt{3} \sec 10^{\circ}$ is equal to:

$=\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}$

$=\frac{\cos 10^{\circ}-\sqrt{3} \sin 10^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$

$=4\left[\frac{\frac{1}{2} \cos 10^{\circ}-\frac{\sqrt{3}}{2} \sin 10^{\circ}}{2 \sin 10^{\circ} \cos 10^{\circ}}\right]$

$=4\left[\frac{\sin \left(30^{\circ}-10^{\circ}\right)}{\sin 20^{\circ}}\right]$ $=4$

Answer key and solution verified before publishing.

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