tan ^{-1}left(tan frac{5 pi}{6}right)+cos ^{- | vedclass

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(A) We know that the principal value branch of $	an^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$ and for $\cos^{-1} x$ is $[0, \pi]$.First,evaluate $

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$0$

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(A) We know that the principal value branch of $	an^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$ and for $\cos^{-1} x$ is $[0, \pi]$.First,evaluate $	an^{-1}(	an \frac{5\pi}{6})$:$	an^{-1}(	an(\pi - \frac{\pi}{6})) = 	an^{-1}(-	an \frac{\pi}{6}) = 	an^{-1}(	an(-\frac{\pi}{6})) = -\frac{\pi}{6}$.Next,evaluate $\cos^{-1}(\cos \frac{13\pi}{6})$:$\cos^{-1}(\cos(2\pi + \frac{\pi}{6})) = \cos^{-1}(\cos \frac{\pi}{6}) = \frac{\pi}{6}$.Adding these results:$-\frac{\pi}{6} + \frac{\pi}{6} = 0$.

$\tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right) = $

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