A) 0
B) 1
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{6}\]
E) \[\frac{\pi }{8}\]
Correct Answer: C
Solution :
\[{{\tan }^{-1}}\left( \frac{c}{a+b} \right)+{{\tan }^{-1}}\left( \frac{b}{a+c} \right)\] \[={{\tan }^{-1}}\left[ \frac{\frac{c}{a+b}+\frac{b}{a+c}}{1-\left( \frac{c}{a+b} \right)\left( \frac{b}{a+c} \right)} \right]\] \[={{\tan }^{-1}}\left[ \frac{ac+{{c}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}+ac+ab+bc-bc} \right]\] \[={{\tan }^{-1}}\left[ \frac{{{a}^{2}}+ac+ab}{{{a}^{2}}+ac+ab} \right]\] \[(\because {{a}^{2}}={{b}^{2}}+{{c}^{2}})\] \[={{\tan }^{-1}}(1)\] \[=\frac{\pi }{4}\]
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