If $\angle A={90}^0$  in $\Delta \text{ABC}$, then ${\tan }^{-1}\left(\frac{c}{a+b}\right)+{\tan }^{-1}\left(\frac{b}{a+c}\right)$  is equal to

We have  
${\tan }^{-1}\left[\frac{c}{a+b}\right]+{\tan }^{-1}\left[\frac{b}{a+c}\right]={\tan }^{-1}\left[\frac{\frac{c}{a+b}+\frac{b}{a+c}}{1-\left(\frac{c}{a+b}\right)\left(\frac{b}{a+c}\right)}\right]={\tan }^{-1}\left[\frac{ac+c^2+ab+b^2}{a^2+ac+ab+bc-bc}\right]$$={\tan }^{-1}\left[\frac{a^2+ac+ab}{a^2+ac+ab}\right][\therefore a^2=b^2+c^2]=\frac{\pi }{4}$