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Question

$\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=$

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Answer

$(d) :$ Let $\cos ^{-1} \frac{\sqrt{3}}{2}=\theta $
$\Rightarrow \cos \theta=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6}$
$\Rightarrow \theta=\frac{\pi}{6} \in[0, \pi]$
$\therefore \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}$

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