MCQ
$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=$
  • A
    $\frac{7 \pi}{6}$
  • $\frac{5 \pi}{6}$
  • C
    $\frac{\pi}{6}$
  • D
    $\frac{3 \pi}{2}$

Answer

Correct option: B.
$\frac{5 \pi}{6}$
Here, we will use $\cos (\pi+x)=\cos (\pi-x)$
$\therefore \cos \left(\frac{7 \pi}{6}\right)=\cos \left(\pi+\frac{\pi}{6}\right)=\cos \left(\pi-\frac{\pi}{6}\right) \\
=\cos \left(\frac{5 \pi}{6}\right) \\
\therefore \cos ^{-1}\left(\cos \left(\frac{7 \pi}{6}\right)\right)=\cos ^{-1}\left(\cos \left(\frac{5 \pi}{6}\right)\right)$
$=\frac{5 \pi}{6}$, which is the required principal value as it is between 0 and $\pi$.

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