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Mathematics · 2023 · 1 mark
CBSE 2023 · Region 1 · Set 1 · Q9
[3/4 tan^2 30^°- sec^2 45^°+ sin^2 60^°] is equal to
$\displaystyle \left[\frac{3}{4} \tan ^{2} 30^{\circ}-\sec ^{2} 45^{\circ}+\sin ^{2} 60^{\circ}\right]$ is equal to
Official answer
(A)
$\displaystyle -1$
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CBSE Class 10 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.