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Mathematics · 2026 · 2 marks
Inverse Trigonometric FunctionsProperties of Inverse Trigonometric Functions2 marksApplyvery short answer
CBSE 2026 · Region 1 · Set 3 · Q23
(a) Simplify: sin^-1 √((1+ cos 2 x)/2), 0<x<(π)/2.
(a)
Simplify : $\displaystyle \sin ^{-1} \sqrt{\frac{1+\cos 2 x}{2}}, 0<x<\frac{\pi}{2}$.
OR
(b)
Evaluate : $\displaystyle \cos \left[\sin ^{-1}(-1)-\tan ^{-1}(-\sqrt{3})\right]$.
Marking-scheme solution
(a)
Given expression $\displaystyle = \sin^{-1}\left(\sqrt{\dfrac{2\cos^2 x}{2}}\right)$
$\displaystyle = \sin^{-1}(\cos x)$
$\displaystyle = \sin^{-1}\left(\sin\left(\dfrac{\pi}{2}-x\right)\right) = \dfrac{\pi}{2}-x$
OR
(b)
Given expression $\displaystyle = \cos\left(-\dfrac{\pi}{2}+\dfrac{\pi}{3}\right)$
$\displaystyle = \cos\left(-\dfrac{\pi}{6}\right) = \dfrac{\sqrt3}{2}$
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CBSE Class 12 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.