cos ^{-1}left(cos frac{13 pi}{6}right)+tan ^{ | vedclass

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

Vedclass · Accepted Answer

$\frac{\pi}{3}$

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(B) We know that the principal value branch of $\cos ^{-1} x$ is $[0, \pi]$ and $	an ^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.First,consider $\cos ^{-1}\left(\cos \frac{13 \pi}{6}ight)$.Since $\frac{13 \pi}{6} = 2 \pi + \frac{\pi}{6}$,we have $\cos \frac{13 \pi}{6} = \cos \frac{\pi}{6}$.Thus,$\cos ^{-1}\left(\cos \frac{13 \pi}{6}ight) = \cos ^{-1}\left(\cos \frac{\pi}{6}ight) = \frac{\pi}{6}$.Next,consider $	an ^{-1}\left(	an \frac{7 \pi}{6}ight)$.Since $\frac{7 \pi}{6} = \pi + \frac{\pi}{6}$,we have $	an \frac{7 \pi}{6} = 	an \frac{\pi}{6}$.Thus,$	an ^{-1}\left(	an \frac{7 \pi}{6}ight) = 	an ^{-1}\left(	an \frac{\pi}{6}ight) = \frac{\pi}{6}$.Adding these results,we get $\frac{\pi}{6} + \frac{\pi}{6} = \frac{2 \pi}{6} = \frac{\pi}{3}$.

$\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)+\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)=$ . . . . . . .

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