Find the principal value of tan ^{-1}(-1). | vedclass

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(D) Let $	an ^{-1}(-1) = y$.Then,$	an y = -1$.Since $	an \left(\frac{\pi}{4}ight) = 1$,we have $	an y = -	an \left(\frac{\pi}{4}ight) = 	an

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$-\frac{\pi}{4}$

Vedclass · Answer

(D) Let $	an ^{-1}(-1) = y$.Then,$	an y = -1$.Since $	an \left(\frac{\pi}{4}ight) = 1$,we have $	an y = -	an \left(\frac{\pi}{4}ight) = 	an \left(-\frac{\pi}{4}ight)$.The range of the principal value branch of $	an ^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$.Since $-\frac{\pi}{4} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$,the principal value of $	an ^{-1}(-1)$ is $-\frac{\pi}{4}$.

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

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