question_answer

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) \[\pi /8\]

Correct Answer: C

Solution :

  \[\angle A={{90}^{o}}\] \[{{\tan }^{-1}}\left( \frac{c}{a+b} \right)+{{\tan }^{-1}}\left( \frac{b}{a+c} \right)\] \[={{\tan }^{-1}}\left[ \frac{\frac{c}{a+b}+\frac{b}{a+c}}{1-\left( \frac{c}{a+b} \right)\left( \frac{b}{a+c} \right)} \right]\] \[={{\tan }^{-1}}\left[ \frac{ca+{{c}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}+ab+ca+bc-bc} \right]\] \[={{\tan }^{-1}}\left[ \frac{{{a}^{2}}+ab+ca}{{{a}^{2}}+ab+ca} \right]\]\[={{\tan }^{-1}}(1)=\frac{\pi }{4}\].