CBSE 2024 · Region 5 · Set 1 · Q21 · 2 marks
Find value of k if \[\sin ^{-1}\left[\mathrm{k} \tan \left(2 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right]=\frac{\pi}{3} \]
Marking-scheme solution
$$\mathrm{k} \tan \left(2 \cos ^{-1} \frac{\sqrt{3}}{2}\right)=\sin \frac{\pi}{3}$\displaystyle \Rightarrow \mathrm{ktan}\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$
$\displaystyle \Rightarrow \mathrm{k} \sqrt{3}=\frac{\sqrt{3}}{2} \quad \therefore \mathrm{k}=\frac{1}{2}$
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.