A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{4}\]
D) none of these
Correct Answer: A
Solution :
| let \[I=\int\limits_{{{\tan }^{-1}}\alpha }^{{{\cot }^{-1}}\alpha }{\frac{\tan x}{\tan x+\cot x}dx}\]\[=\int\limits_{{{\tan }^{-1}}\alpha }^{{{\cot }^{-1}}\alpha }{\frac{\cot x}{\cot x+\tan x}dx}\] \[\left( \because {{\tan }^{-1}}\alpha +{{\cot }^{-1}}\alpha =\frac{\pi }{2} \right)\] |
| \[\therefore \] \[2I=\int\limits_{{{\tan }^{-1}}\alpha }^{{{\cot }^{-1}}\alpha }{dx={{\cot }^{-1}}\alpha -{{\tan }^{-1}}\alpha }\] \[\Rightarrow \] \[I=\frac{1}{2}\left[ \frac{\pi }{2}-2{{\tan }^{-1}}\alpha \right]\] |
| \[\therefore \] \[-\frac{\pi }{2}<I<\frac{3\pi }{4}\forall \alpha \] |
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