Explanation

(c)
$\because\ \ \mathrm{\Delta}ABC$ is right angled at $A$.
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a^{2} = b^{2} + c^{2}$ ...(i)
Now, $\tan^{- 1}{\left( \frac{c}{a + b} \right) + \tan^{- 1}\left( \frac{b}{a + c} \right)}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}\left\lbrack \frac{\frac{c}{a + b} + \frac{b}{a + c}}{1 - \left( \frac{c}{a + b} \right)\left( \frac{b}{a + c} \right)} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}\left\lbrack \frac{ac + c^{2} + ab + b^{2}}{a^{2} + ac + ab + bc - bc} \right\rbrack$
${\ \ \ \ \ \ \ \ \ \ \ = tan}^{- 1}\left\lbrack \frac{a^{2} + ac + ab}{a^{2} + ac + ab} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}{(1) = \frac{\pi}{4}}$ [using Eq. (i)]