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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & 	an ^{-1}\left(	an \frac{5 \pi}{6}ight)+\cos ^{-1}\left(\cos \frac{13 \pi}{6}ight) \ & =	an ^{-1}\left[	an \left(\pi-\frac{\pi}{6}ight)ight]+\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{6}ight)ight] \ & =	an ^{-1}\left[-	an \frac{\pi}{6}ight]+\cos ^{-1}\left[\cos \left(\frac{\pi}{6}ight)ight] \ & =	an ^{-1}\left[	an \left(-\frac{\pi}{6}ight)ight]+\cos ^{-1}\left[\cos \left(\frac{\pi}{6}ight)ight]=-\frac{\pi}{6}+\frac{\pi}{6}=0\end{aligned}$

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

Solution