$\sin ^{-1}\left[\sin \left(-600^{\circ}\right)\right]+\cot ^{-1}(-\sqrt{3})=$

Question

Aicharya · Accepted Answer

$\begin{aligned} & \sin ^{-1}\left[\sin (-600)^{\circ}ight]+\cot ^{-1}(\sqrt{3}) \ & =\sin ^{-1}\left[-\sin \left(360^{\circ}+180^{\circ}+60^{\circ}ight)ight]+\left[-\cot ^{-1}(\sqrt{3})ight] \ & =\sin ^{-1}\left[\sin \left(60^{\circ}ight)ight]+\left[-	an ^{-1}\left(\frac{1}{\sqrt{3}}ight)ight] \ & =\sin ^{-1}\left[\sin 60^{\circ}ight]-\frac{\pi}{6}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6} \end{aligned}$

$\sin ^{-1}\left[\sin \left(-600^{\circ}\right)\right]+\cot ^{-1}(-\sqrt{3})=$

Solution