A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{8}\]
E) \[\frac{2\pi }{3}\]
Correct Answer: A
Solution :
\[{{\tan }^{-1}}\left( \frac{m}{n} \right)-{{\tan }^{-1}}\left( \frac{m-n}{m+n} \right)\] \[={{\tan }^{-1}}\left\{ \frac{\frac{m}{n}-\frac{m-n}{m+n}}{1+\left( \frac{m}{n} \right)\left( \frac{m-n}{m+n} \right)} \right\}\] \[={{\tan }^{-1}}\left\{ \frac{{{m}^{2}}+mn-mn+{{n}^{2}}}{mn+{{n}^{2}}+{{m}^{2}}-mn} \right\}\] \[={{\tan }^{-1}}\left\{ \frac{{{m}^{2}}+{{n}^{2}}}{{{m}^{2}}+{{n}^{2}}} \right\}=\frac{\pi }{4}\]
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