# 19. $\cos^{-1}\left(\cos\frac{7\pi}{6} \right )$ is equal to         (A)    $\frac{7\pi}{6}$        (B)    $\frac{5\pi}{6}$        (C)    $\frac{\pi}{3}$        (D)    $\frac{\pi}{6}$

D Divya Prakash Singh

As we know that $\cos^{-1} (cos x ) = x$ if $x\epsilon [0,\pi]$ and is principal value range of $\cos^{-1}x$.

In this case $\cos^{-1}\left(\cos\frac{7\pi}{6} \right )$,

$\frac{7\pi}{6} \notin [0,\pi]$

hence we have then,

$\cos^{-1}\left(\cos\frac{7\pi}{6} \right ) =$  $\cos^{-1} \left ( \cos \frac{-7\pi}{6} \right ) = \cos^{-1}\left [ \cos\left ( 2\pi - \frac{7\pi}{6} \right ) \right ]$

$\left [ \because \cos (2\pi + x) = \cos x \right ]$

$\therefore\ we\ have \cos^{-1}\left ( \cos \frac{7\pi}{6} \right ) = \cos^{-1}\left ( \cos \frac{5\pi}{6} \right ) = \frac{5\pi}{6}$

Hence the correct answer is $\frac{5\pi}{6}$ (B).

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