Find the principal value of tan ^{-1}(-sqrt{3 | vedclass

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(B) Let $	an ^{-1}(-\sqrt{3}) = y$. Then,$	an y = -\sqrt{3}$. Since $	an \frac{\pi}{3} = \sqrt{3}$,we have $	an y = -	an \frac{\pi}{3} = 	an \le

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$-\frac{\pi}{3}$

Vedclass · Answer

(B) Let $	an ^{-1}(-\sqrt{3}) = y$. Then,$	an y = -\sqrt{3}$. Since $	an \frac{\pi}{3} = \sqrt{3}$,we have $	an y = -	an \frac{\pi}{3} = 	an \left(-\frac{\pi}{3}ight)$. The range of the principal value branch of $	an ^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$. Since $-\frac{\pi}{3} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$,the principal value of $	an ^{-1}(-\sqrt{3})$ is $-\frac{\pi}{3}$.

Find the principal value of $\tan ^{-1}(-\sqrt{3})$

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