sin ^{-1}left(-frac{1}{2}right)+cos ^{-1}left | vedclass

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(D) We know that $\sin ^{-1}(-x) = -\sin ^{-1}(x)$ for $x \in [-1, 1]$.Therefore,$\sin ^{-1}\left(-\frac{1}{2}\right) = -\sin ^{-1}\left(\frac{1}{2}\r

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) We know that $\sin ^{-1}(-x) = -\sin ^{-1}(x)$ for $x \in [-1, 1]$.Therefore,$\sin ^{-1}\left(-\frac{1}{2}\right) = -\sin ^{-1}\left(\frac{1}{2}\right) = -\frac{\pi}{6}$.We also know that $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$ because $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.Adding these values,we get: $-\frac{\pi}{6} + \frac{\pi}{6} = 0$.Thus,the correct option is $D$.

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

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