A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{4}\]or \[-\frac{3\pi }{4}\]
Correct Answer: C
Solution :
\[{{\tan }^{-1}}\frac{x}{y}-{{\tan }^{-1}}\left( \frac{x-y}{x+y} \right)={{\tan }^{-1}}\frac{x}{y}-{{\tan }^{-1}}\left( \frac{1-y/x}{1+y/x} \right)\] \[={{\tan }^{-1}}\frac{x}{y}-\left( {{\tan }^{-1}}1-{{\tan }^{-1}}\frac{y}{x} \right)\] \[={{\tan }^{-1}}\frac{x}{y}+{{\tan }^{-1}}\frac{y}{x}-\frac{\pi }{4}\] \[={{\tan }^{-1}}\frac{x}{y}+{{\cot }^{-1}}\frac{x}{y}-\frac{\pi }{4}=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}\].
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