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

Inverse Trigonometric Functions

Solution:

$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=\cos ^{-1}\left[\cos \left(2 \pi-\frac{5 \pi}{6}\right)\right]$
where, $\frac{5 \pi}{6} \in[0, \pi]$
$\Rightarrow \cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=\cos ^{-1}\left[ \cos \left(\frac{5 \pi}{6}\right)\right]$
$[\because \cos (2 \pi-\theta)=\cos \theta]$