A) 0
B) 1
C) \[\pi /4\]
D) \[\pi /6\]
E) \[\pi /8\]
Correct Answer: C
Solution :
\[\angle A={{90}^{o}}\] \[{{\tan }^{-1}}\left( \frac{c}{a+b} \right)+{{\tan }^{-1}}\left( \frac{b}{a+c} \right)\] \[={{\tan }^{-1}}\left[ \frac{\frac{c}{a+b}+\frac{b}{a+c}}{1-\left( \frac{c}{a+b} \right)\left( \frac{b}{a+c} \right)} \right]\] \[={{\tan }^{-1}}\left[ \frac{ca+{{c}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}+ab+ca+bc-bc} \right]\] \[={{\tan }^{-1}}\left[ \frac{{{a}^{2}}+ab+ca}{{{a}^{2}}+ab+ca} \right]\]\[={{\tan }^{-1}}(1)=\frac{\pi }{4}\].You need to login to perform this action.
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