If $f : R \to R$ be given by $f(x) = \tan x$, then $f^{-1}(1)$ is

$\frac{\pi}{4}$

The correct answer is Option (1) → $\frac{\pi}{4}$ ##

Given that, $f(x) = \tan x$

Let $y = \tan x$

$\Rightarrow x = \tan^{-1} y$

$\Rightarrow f^{-1}(x) = \tan^{-1} x$

$\Rightarrow f^{-1}(1) = \tan^{-1} 1$

$= \tan^{-1} \left( \tan \frac{\pi}{4} \right) = \frac{\pi}{4}$   $\left[ ∵\tan \frac{\pi}{4} = 1 \right]$

$[∵\tan^{-1}(\tan \theta) = \theta]$