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}

Answers (1)

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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