A) \[\frac{3\pi }{4}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{3}\]
D) \[{{\tan }^{-1}}(6)\]
E) \[{{\tan }^{-1}}(5)\]
Correct Answer: A
Solution :
\[{{\tan }^{-1}}(2)+{{\tan }^{-1}}(3)\] \[={{\tan }^{-1}}\left( \frac{2+3}{1-2.3} \right)={{\tan }^{-1}}\left( \frac{5}{1-6} \right)\] \[\left[ \because {{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right) \right]\] \[={{\tan }^{-1}}\left( \frac{5}{-5} \right)={{\tan }^{-1}}(-1)\] \[={{\tan }^{-1}}\left( \tan \left( \frac{\pi }{2}+\frac{\pi }{4} \right) \right)=\frac{3\pi }{4}\]You need to login to perform this action.
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