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Q. The principal value of $\cos ^{-1}\left(-\sin \frac{7 \pi}{6}\right)$, is

Inverse Trigonometric Functions

Solution:

We have.
$\cos ^{-1}\left(-\sin \frac{7 \pi}{6}\right)=\cos ^{-1}\left(-\sin \left(\pi+\frac{\pi}{6}\right)\right)$
$=\cos ^{-1}\left(\sin \frac{\pi}{6}\right) $
$=\frac{\pi}{2}-\sin ^{-1}\left(\sin \frac{\pi}{6}\right)=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}$