The smallest and the largest values of ${\tan }^{-1}\left(\frac{1-x}{1+x}\right),0\le x\le 1$ are

We have,

   ${\tan }^{-1}\left(\frac{1-x}{1+x}\right)={\tan }^{-1}1-{\tan }^{-1}x=\frac{\pi }{4}-{\tan }^{-1}x$

We have,  $0\le x\le 1$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\le {\tan }^{-1}x\le \frac{\pi }{4}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\ge -{\tan }^{-1}x\ge -\frac{\pi }{4}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{\pi }{4}\ge \frac{\pi }{4}-{\tan }^{-1}x\ge 0\Rightarrow \frac{\pi }{4}\ge {\tan }^{-1}\left(\frac{1-x}{1+\frac{x}{x}}\right)\ge 0\end{array}$