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Mathematics · 2026 · 2 marks
Inverse Trigonometric FunctionsProperties of Inverse Trigonometric Functions2 marksApplyvery short answer
CBSE 2026 · Region 1 · Set 1 · Q25
(a) Simplify: tan^-1((cos 2 x- sin 2 x)/(cos 2 x+ sin 2 x)), 0<x<(π)/4.
(a)
Simplify : $\displaystyle \tan ^{-1}\left(\frac{\cos 2 x-\sin 2 x}{\cos 2 x+\sin 2 x}\right), 0<x<\frac{\pi}{4}$.
OR
(b)
Evaluate : $\displaystyle \tan \left(\sin ^{-1} 1-\cos ^{-1}\left(-\frac{1}{2}\right)\right)$
Official answer
(a)
π/$\displaystyle 4$ - x; (b) √$\displaystyle 3$
Marking-scheme solution
(a)
$\displaystyle \tan^{-1}\left(\dfrac{\cos2x-\sin2x}{\cos2x+\sin2x}\right) = \tan^{-1}\left(\dfrac{1-\tan2x}{1+\tan2x}\right)$
$\displaystyle = \tan^{-1}\left[\tan\left(\dfrac{\pi}{4}-2x\right)\right] = \dfrac{\pi}{4}-2x$
OR
(b)
$\displaystyle \tan\left(\sin^{-1}1-\cos^{-1}\left(-\dfrac12\right)\right) = \tan\left(\dfrac{\pi}{2}-\dfrac{2\pi}{3}\right)$
$\displaystyle = \tan\left(-\dfrac{\pi}{6}\right) = -\dfrac{1}{\sqrt3}$
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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.