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Inverse Trigonometric Functions — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSInverse Trigonometric Functions1 Mark
Question
The principal solution of $\sin ^{-1}\left(\sin \left(\frac{5 \pi}{3}\right)\right)$ is
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Answer

$\text { (d) : In }\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
$\sin ^{-1}\left(\sin \left(\frac{5 \pi}{3}\right)\right)=\sin ^{-1}\left(\sin \left(2 \pi-\frac{\pi}{3}\right)\right)$
$=\sin ^{-1}\left(\sin \left(\frac{-\pi}{3}\right)\right)=\frac{-\pi}{3}$

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