The value of sin ^{-1}left(-frac{1}{2}right)+ | vedclass

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(C) Let $\alpha = \sin ^{-1}\left(-\frac{1}{2}\right)$. Since the range of $\sin ^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,we have $\si

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$\cos ^{-1}\left(-\frac{1}{2}\right)$

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(C) Let $\alpha = \sin ^{-1}\left(-\frac{1}{2}\right)$. Since the range of $\sin ^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,we have $\sin \alpha = -\frac{1}{2} = \sin\left(-\frac{\pi}{6}\right)$,so $\alpha = -\frac{\pi}{6}$.Let $\beta = \cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$. Since the range of $\cos ^{-1} x$ is $[0, \pi]$,we have $\cos \beta = -\frac{\sqrt{3}}{2} = -\cos\left(\frac{\pi}{6}\right) = \cos\left(\pi - \frac{\pi}{6}\right) = \cos\left(\frac{5\pi}{6}\right)$,so $\beta = \frac{5\pi}{6}$.Adding these values,we get $\alpha + \beta = -\frac{\pi}{6} + \frac{5\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$.Now,checking the options:Option $C$ is $\cos ^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$.Thus,the correct option is $C$.

The value of $\sin ^{-1}\left(-\frac{1}{2}\right)+\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ is

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