In a ∆ ABC, if C is a right angle, then

(b) $\frac{\pi}{4}$

We know

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

$\therefore \tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)=\tan ^{-1}\left(\frac{\frac{a}{b+c}+\frac{b}{c+a}}{1-\frac{a}{b+c} \times \frac{b}{c+a}}\right)$

$=\tan ^{-1}\left(\frac{\frac{a c+a^{2}+b^{2}+b c}{(b+c)(c+a)}}{\frac{a c+c^{2}+b c}{(b+c)(c+a)}}\right)$

$=\tan ^{-1}\left(\frac{a c+c^{2}+b c}{a c+c^{2}+b c}\right) \quad\left[\because a^{2}+b^{2}=c^{2}\right]$

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

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

$=\frac{\pi}{4}$