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Mathematics · 2026 · 1 mark
CBSE 2026 · Region 3 · Set 2 · Q11
The value of (1/3 cot^2 30^°-1/2 sec^2 60^°) is:
The value of $\displaystyle \left(\frac{1}{3} \cot ^{2} 30^{\circ}-\frac{1}{2} \sec ^{2} 60^{\circ}\right)$ is :
Official answer
(A)
$\displaystyle - 1$
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CBSE Class 10 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.