In a $$\vartriangle ABC,A = {90^ \circ }.$$    Then $${\tan^{ - 1}}\frac{b}{{a + c}} + {\tan ^{ - 1}}\frac{c}{{a + b}}$$      is equal to

Solution :
$$\eqalign{ & {\tan ^{ - 1}}\left( {\frac{a}{{b + c}}} \right) + {\tan ^{ - 1}}\left( {\frac{b}{{c + a}}} \right) \cr & {\text{We know that }}{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right) \cr & {\text{Replace, }}x{\text{ by }}\frac{a}{{b + c}}{\text{ and }}y{\text{ by }}\frac{b}{{c + a}} \cr & = {\tan ^{ - 1}}\left( {\frac{{\frac{a}{{b + c}} + \frac{b}{{c + a}}}}{{1 - \frac{{ab}}{{\left( {b + c} \right)\left( {c + a} \right)}}}}} \right) \cr & = {\tan ^{ - 1}}\left( {\frac{{\frac{{ac + {a^2} + {b^2} + bc}}{{\left( {b + c} \right)\left( {c + a} \right)}}}}{{\frac{{\left( {b + c} \right)\left( {c + a} \right) - ab}}{{\left( {b + c} \right)\left( {c + a} \right)}}}}} \right) \cr & = {\tan ^{ - 1}}\left( {\frac{{\frac{{{a^2} + {b^2} + bc + ac}}{{\left( {b + c} \right)\left( {c + a} \right)}}}}{{\frac{{bc + {c^2} + ca}}{{\left( {b + c} \right)\left( {c + a} \right)}}}}} \right) \cr & = {\tan ^{ - 1}}\frac{{{a^2} + {b^2} + bc + ac}}{{bc + {c^2} + ca}} \cr & {\text{Given, }}\angle {\text{C = }}{90^ \circ } \Rightarrow {a^2} + {b^2} = {c^2} \cr & = {\tan ^{ - 1}}\frac{{{c^2} + bc + ac}}{{bc + {c^2} + ca}} \cr & = {\tan ^{ - 1}}1 \cr & = \frac{\pi }{4} \cr} $$