sin ^{-1}left(sin frac{2 pi}{3}right)+cos ^{- | vedclass

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(C) We evaluate each term individually using the principal value branches of the inverse trigonometric functions:$1$. For $\sin ^{-1}\left(\sin \frac{

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$\frac{11 \pi}{12}$

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(C) We evaluate each term individually using the principal value branches of the inverse trigonometric functions:$1$. For $\sin ^{-1}\left(\sin \frac{2 \pi}{3}ight)$:Since $\frac{2 \pi}{3}$ is not in the principal range $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we write $\sin \frac{2 \pi}{3} = \sin(\pi - \frac{\pi}{3}) = \sin \frac{\pi}{3}$.Thus,$\sin ^{-1}\left(\sin \frac{\pi}{3}ight) = \frac{\pi}{3}$.$2$. For $\cos ^{-1}\left(\cos \frac{7 \pi}{6}ight)$:Since $\frac{7 \pi}{6}$ is not in the principal range $[0, \pi]$,we write $\cos \frac{7 \pi}{6} = \cos(2\pi - \frac{5\pi}{6}) = \cos \frac{5 \pi}{6}$.Thus,$\cos ^{-1}\left(\cos \frac{5 \pi}{6}ight) = \frac{5 \pi}{6}$.$3$. For $	an ^{-1}\left(	an \frac{3 \pi}{4}ight)$:Since $\frac{3 \pi}{4}$ is not in the principal range $(-\frac{\pi}{2}, \frac{\pi}{2})$,we write $	an \frac{3 \pi}{4} = 	an(\pi - \frac{\pi}{4}) = -	an \frac{\pi}{4} = 	an(-\frac{\pi}{4})$.Thus,$	an ^{-1}\left(	an(-\frac{\pi}{4})ight) = -\frac{\pi}{4}$.Summing these values:$\frac{\pi}{3} + \frac{5 \pi}{6} - \frac{\pi}{4} = \frac{4\pi + 10\pi - 3\pi}{12} = \frac{11 \pi}{12}$.

$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)+\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)+\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)$ is equal to

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