tan ^{-1}left[frac{1}{sqrt{3}} sin frac{5 pi} | vedclass

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(C) Let the expression be $E = 	an ^{-1}\left[\frac{1}{\sqrt{3}} \sin \frac{5 \pi}{2}ight] + \sin ^{-1}\left[\cos \left(\sin ^{-1} \frac{\sqrt{3}}{

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(C) Let the expression be $E = 	an ^{-1}\left[\frac{1}{\sqrt{3}} \sin \frac{5 \pi}{2}ight] + \sin ^{-1}\left[\cos \left(\sin ^{-1} \frac{\sqrt{3}}{2}ight)ight]$.First,evaluate $\sin \frac{5 \pi}{2} = \sin \left(2 \pi + \frac{\pi}{2}ight) = \sin \frac{\pi}{2} = 1$.Next,evaluate $\sin ^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{3}$.Now substitute these values into the expression:$E = 	an ^{-1}\left[\frac{1}{\sqrt{3}} 	imes 1ight] + \sin ^{-1}\left[\cos \frac{\pi}{3}ight]$.Since $\cos \frac{\pi}{3} = \frac{1}{2}$,we have:$E = 	an ^{-1}\left(\frac{1}{\sqrt{3}}ight) + \sin ^{-1}\left(\frac{1}{2}ight)$.We know that $	an ^{-1}\left(\frac{1}{\sqrt{3}}ight) = \frac{\pi}{6}$ and $\sin ^{-1}\left(\frac{1}{2}ight) = \frac{\pi}{6}$.Therefore,$E = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2 \pi}{6} = \frac{\pi}{3}$.

$\tan ^{-1}\left[\frac{1}{\sqrt{3}} \sin \frac{5 \pi}{2}\right] + \sin ^{-1}\left[\cos \left(\sin ^{-1} \frac{\sqrt{3}}{2}\right)\right]$ is equal to

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