A) \[2\pi +a\]
B) \[\pi +a\]
C) 0
D) \[\pi -a\]
Correct Answer: C
Solution :
[c] For \[\alpha \in \left( -\frac{3\pi }{2},-\pi \right),\tan \alpha <0\] \[\Rightarrow {{\tan }^{-1}}(cot\alpha )-co{{t}^{-1}}(tan\alpha )\times \] \[{{\tan }^{-1}}(cot\alpha )-\left[ \frac{\pi }{2}-{{\tan }^{-1}}(tan\alpha ) \right]\] \[={{\tan }^{-1}}(cot\alpha )+{{\tan }^{-1}}(tan\alpha )-\frac{\pi }{2}\] \[=-\pi \] Also for points in the second quadrant, we have \[{{\sin }^{-1}}(\sin \alpha )+{{\cos }^{-1}}(cos\alpha )=\pi \]
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