A) \[{{\tan }^{-1}}\frac{n}{m}\]
B) \[{{\tan }^{-1}}\frac{m+n}{m-n}\]
C) \[\frac{\pi }{4}\]
D) \[{{\tan }^{-1}}\left( \frac{1}{2} \right)\]
E) \[\frac{\pi }{2}\]
Correct Answer: C
Solution :
\[{{\tan }^{-1}}\frac{m}{n}-{{\tan }^{-1}}\frac{m-n}{m+n}\] \[={{\tan }^{-1}}\frac{m}{n}-{{\tan }^{-1}}\frac{\frac{m}{n}-1}{1+\frac{m}{n}}\] \[={{\tan }^{-1}}\frac{m}{n}-{{\tan }^{-1}}\frac{m}{n}+{{\tan }^{-1}}(1)\] \[={{\tan }^{-1}}\left( \tan \frac{\pi }{4} \right)=\frac{\pi }{4}\]You need to login to perform this action.
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