If \[\angle A={{90}^{o}}\] in the triangle ABC, then \[{{\tan }^{-1}}\left( \frac{c}{a+b} \right)+{{\tan }^{-1}}\left( \frac{b}{a+c} \right)=\] [Kerala (Engg.) 2005]

If \[\angle A={{90}^{o}}\] in the triangle ABC, then \[{{\tan }^{-1}}\

1.	If \[\angle A={{90}^{o}}\] in the triangle ABC, then \[{{\tan }^{-1}}\left( \frac{c}{a+b} \right)+{{\tan }^{-1}}\left( \frac{b}{a+c} \right)=\] [Kerala (Engg.) 2005]
A.	0
B.	1
C.	\[\pi /4\]
D.	\[\pi /6\]
E.	\[\pi /8\]
Answer» D. \[\pi /6\]

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If \[\angle A={{90}^{o}}\] in the triangle ABC, then \[{{\tan }^{-1}}\left( \frac{c}{a+b} \right)+{{\tan }^{-1}}\left( \frac{b}{a+c} \right)=\] [Kerala (Engg.) 2005]

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