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A value of $\tan ^{-1}\left(\sin \left(\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)\right)\right)$ is
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Verified Answer
The correct answer is:
$\frac{\pi}{6}$
$\frac{\pi}{6}$
Consider $\tan ^{-1}\left[\sin \left(\cos ^{-1} \sqrt{\frac{2}{3}}\right)\right]$
$$
\begin{aligned}
& \text { Let } \cos ^{-1} \sqrt{\frac{2}{3}}=\theta \Rightarrow \cos \theta=\sqrt{\frac{2}{3}} \\
& \Rightarrow \quad \sin \theta=\sqrt{1-\cos ^2 \theta}=\sqrt{1-\frac{2}{3}}=\sqrt{\frac{1}{3}} \\
& \therefore \quad \tan ^{-1}\left[\sin \left(\cos ^{-1} \sqrt{\frac{2}{3}}\right)\right]=\tan ^{-1}[\sin \theta] \\
& =\tan ^{-1}\left[\sqrt{\frac{1}{3}}\right]=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) \\
& =\frac{\pi}{6}
\end{aligned}
$$
$$
\begin{aligned}
& \text { Let } \cos ^{-1} \sqrt{\frac{2}{3}}=\theta \Rightarrow \cos \theta=\sqrt{\frac{2}{3}} \\
& \Rightarrow \quad \sin \theta=\sqrt{1-\cos ^2 \theta}=\sqrt{1-\frac{2}{3}}=\sqrt{\frac{1}{3}} \\
& \therefore \quad \tan ^{-1}\left[\sin \left(\cos ^{-1} \sqrt{\frac{2}{3}}\right)\right]=\tan ^{-1}[\sin \theta] \\
& =\tan ^{-1}\left[\sqrt{\frac{1}{3}}\right]=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) \\
& =\frac{\pi}{6}
\end{aligned}
$$
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