tan ^{-1}left(tan frac{31 pi}{6}right)= . . | vedclass

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Question

(A) We know that the range of the principal value branch of $	an ^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.Given expression is $	an ^{-1}(	an \

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(A) We know that the range of the principal value branch of $	an ^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.Given expression is $	an ^{-1}(	an \frac{31 \pi}{6})$.First,simplify the angle $\frac{31 \pi}{6}$:$\frac{31 \pi}{6} = \frac{30 \pi + \pi}{6} = 5 \pi + \frac{\pi}{6}$.Since $	an(n \pi + 	heta) = 	an 	heta$,we have $	an(5 \pi + \frac{\pi}{6}) = 	an \frac{\pi}{6}$.Therefore,$	an ^{-1}(	an \frac{31 \pi}{6}) = 	an ^{-1}(	an \frac{\pi}{6})$.Since $\frac{\pi}{6} \in (-\frac{\pi}{2}, \frac{\pi}{2})$,the value is $\frac{\pi}{6}$.Thus,the correct option is $A$.

$\tan ^{-1}\left(\tan \frac{31 \pi}{6}\right)=$ . . . . . . .

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