CBSE 2025 · Region 6 · Set 1 · Q5 · 1 mark
$\displaystyle \left[\sec ^{-1}(-\sqrt{2})-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\right]$ is equal to :
Official answer
From CBSE’s own marking scheme for this paper.
(D)
$\displaystyle \frac{7 \pi}{12}$
Inverse Trigonometric FunctionsProperties of Inverse Trigonometric FunctionsApplymcqmedium
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CBSE Class 12 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.