Question

If $f:(-\frac{\pi }{2},\frac{\pi }{2})\to (-\infty ,\infty )$ is defined by $f\left(x\right)=\tan x$, then $f^{-1}\left(\sqrt{3}\right)=$

Answer 1

Given, $f:(-\frac{\pi }{2},\frac{\pi }{2})\to (-\infty ,\infty )$ is defined by $f\left(x\right)=\tan x$.

This function $f\left(x\right)$ is one-one and onto in the given domain and range.

So it is invertible and and the inverse is defined by $f^{-1}\left(y\right)={\tan }^{-1}y$ for $y\in (-\infty ,\infty )$.

So $f^{-1}\left(\sqrt{3}\right)={\tan }^{-1}\left(\sqrt{3}\right)=\frac{\pi }{3}$. [ Since $\tan {60}^o=\sqrt{3}$].