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Mathematics · 2023 · 1 mark
CBSE 2023 · Region 1 · Set 2 · Q17
[5/8 sec^2 60^°- tan^2 60^°+ cos^2 45^°] is equal to
$\displaystyle \left[\frac{5}{8} \sec ^{2} 60^{\circ}-\tan ^{2} 60^{\circ}+\cos ^{2} 45^{\circ}\right]$ is equal to
Official answer
(C)
$\displaystyle 0$
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