Evaluate: sin ^{-1}left[sin left(-600^{circ}r | vedclass

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(D) We need to evaluate $\sin ^{-1}[\sin (-600^{\circ})] + \cot ^{-1}(-\sqrt{3})$.First,simplify $\sin (-600^{\circ})$:$\sin (-600^{\circ}) = -\sin (6

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$\frac{7 \pi}{6}$

Vedclass · Answer

(D) We need to evaluate $\sin ^{-1}[\sin (-600^{\circ})] + \cot ^{-1}(-\sqrt{3})$.First,simplify $\sin (-600^{\circ})$:$\sin (-600^{\circ}) = -\sin (600^{\circ}) = -\sin (360^{\circ} + 240^{\circ}) = -\sin (240^{\circ}) = -\sin (180^{\circ} + 60^{\circ}) = -(-\sin 60^{\circ}) = \sin 60^{\circ}$.Thus,$\sin ^{-1}[\sin (-600^{\circ})] = \sin ^{-1}(\sin 60^{\circ}) = 60^{\circ} = \frac{\pi}{3}$.Next,evaluate $\cot ^{-1}(-\sqrt{3})$:Since $\cot ^{-1}(-x) = \pi - \cot ^{-1}(x)$,we have $\cot ^{-1}(-\sqrt{3}) = \pi - \cot ^{-1}(\sqrt{3}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.Adding the two results:$\frac{\pi}{3} + \frac{5\pi}{6} = \frac{2\pi + 5\pi}{6} = \frac{7\pi}{6}$.

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

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