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Q. $\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$ is equal to

Inverse Trigonometric Functions

Solution:

$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$
$=\pi-\cos ^{-1} \frac{\sqrt{3}}{2}=\pi-\frac{\pi}{6}=\frac{5 \pi}{6}$.