Mathematics · 2026
JEE Main · 24 January 2026, Shift 1 · Q14
The value of (√ 3 cosec 20^°- sec 20^°)/(cos 20^° cos 40^° cos 60^° cos 80^°) is equal to
The value of $\displaystyle \frac{\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}}{\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}}$ is equal to
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 64$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.