CBSE 2024 · Region 3 · Set 2 · Q24 · 2 marks
Find the value of $\displaystyle \left[\sin ^{2}\left\{\cos ^{-1}\left(\frac{3}{5}\right)\right\}+\tan ^{2}\left\{\sec ^{-1}(3)\right\}\right]$.
Marking-scheme solution
Required value $\displaystyle =\left[1-\cos ^{2}\left(\cos ^{-1} \frac{3}{5}\right)\right]+\left[\sec ^{2}\left(\sec ^{-1} 3\right)-1\right]$=\left(1-\frac{9}{25}\right)+(9-1)$\displaystyle =\frac{216}{25}$
Inverse Trigonometric FunctionsProperties of Inverse Trigonometric FunctionsApplyvery_short_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.