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Mathematics · 2025 · 2 marks
Inverse Trigonometric FunctionsProperties of Inverse Trigonometric Functions2 marksApplyvery short answer
CBSE 2025 · Region 1 · Set 1 · Q22
Evaluate: tan^-1[2 sin (2 cos^-1 (√ 3)/2)]
Evaluate $\displaystyle : \tan ^{-1}\left[2 \sin \left(2 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right]$
Marking-scheme solution
\[\begin{aligned}
& \tan ^{-1}\left[2 \sin \left(2 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right] \\
& =\tan ^{-1}\left[2 \sin \left(2 \times \frac{\pi}{6}\right)\right]=\tan ^{-1}\left[2 \sin \frac{\pi}{3}\right] \\
& =\tan ^{-1}\left[2 \times \frac{\sqrt{3}}{2}\right]=\tan ^{-1} \sqrt{3}=\frac{\pi}{3}
\end{aligned}
\]
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CBSE Class 12 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.