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

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Aicharya · Accepted Answer

Let $\sin ^{-1}\left(\frac{-1}{2}ight)=\alpha$, where $\frac{-\pi}{2} \leq \alpha \leq \frac{\pi}{2}$ $\sin \alpha=\frac{-1}{2}=-\sin \frac{\pi}{6}=\sin \left(\frac{-\pi}{6}ight) \Rightarrow \alpha=-\pi / 6$ Let $\cos ^{-1}\left(\frac{-\sqrt{3}}{2}ight)=\beta$, where $0 \leq \beta \leq \pi$ $\cos \beta=\frac{-\sqrt{3}}{2}=\frac{-\cos \pi}{6} \Rightarrow \cos \left(\pi-\frac{\pi}{6}ight)=\frac{\cos 5 \pi}{6} \Rightarrow \beta=\frac{5 \pi}{6}$ $	herefore \sin ^{-1}\left(\frac{-1}{2}ight)+\cos ^{-1}\left(\frac{-\sqrt{3}}{2}ight)=\frac{-\pi}{6}+\frac{5 \pi}{6} \quad=\frac{4 \pi}{6}=\frac{2 \pi}{3}=\cos ^{-1}\left(\frac{-1}{2}ight)$

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

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