A value of tan^{-1} left( sin left( cos^{-1} | vedclass

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(D) Let $	heta = \cos^{-1} \sqrt{\frac{2}{3}}$.Then $\cos 	heta = \sqrt{\frac{2}{3}}$.We know that $\sin 	heta = \sqrt{1 - \cos^2 	heta}$.Substitu

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$\frac{\pi}{6}$

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(D) Let $	heta = \cos^{-1} \sqrt{\frac{2}{3}}$.Then $\cos 	heta = \sqrt{\frac{2}{3}}$.We know that $\sin 	heta = \sqrt{1 - \cos^2 	heta}$.Substituting the value of $\cos 	heta$,we get $\sin 	heta = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$.Now,the expression becomes $	an^{-1} (\sin 	heta) = 	an^{-1} \left( \frac{1}{\sqrt{3}} ight)$.Since $	an \left( \frac{\pi}{6} ight) = \frac{1}{\sqrt{3}}$,we have $	an^{-1} \left( \frac{1}{\sqrt{3}} ight) = \frac{\pi}{6}$.

$A$ value of $\tan^{-1} \left( \sin \left( \cos^{-1} \sqrt{\frac{2}{3}} \right) \right)$ is

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