A) 0
B) \[\frac{\pi }{2}\]
C) \[\pi \]
D) \[\frac{3\pi }{2}\]
Correct Answer: C
Solution :
| \[D>0;\] \[\alpha +\beta <0\Rightarrow \alpha <0;\] \[\beta <0\] |
| \[\therefore {{\cot }^{-1}}\alpha +{{\cot }^{-1}}\left( \frac{1}{\alpha } \right)-\frac{\pi }{2}\] |
| \[={{\cot }^{-1}}\alpha +\pi +{{\tan }^{-1}}\alpha -\frac{\pi }{2}\] |
| \[=\pi +\frac{\pi }{2}-\frac{\pi }{2}=\pi \] |
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