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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]2 Marks
Question
Evaluate $\cos \left[\frac{\pi}{6}+\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]$
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Answer

$ \text { Let } \cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)=y$
$\therefore \cos y=\frac{-\sqrt{3}}{2}$
$=-\cos \left(\frac{\pi}{6}\right)$
$=\cos \left(\pi-\frac{\pi}{6}\right)$
$=\cos \frac{5 \pi}{6} $
The principal value branch of $\cos ^{-1}$ is $[0, \pi]$ and $0 \leq \frac{5 \pi}{6} \leq \pi$.
$ \therefore y=\frac{5 \pi}{6}$
$\therefore \cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)=\frac{5 \pi}{6}$
$\therefore \frac{\pi}{6}+\cos ^{-1}\left(\frac{-\sqrt{3}}{3}\right)$
$=\frac{\pi}{6}+\frac{5 \pi}{6}$
$=\pi$
$\therefore \cos \left[\frac{\pi}{6}+\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]=\cos \pi=-1 $

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