A) \[\frac{2}{\sin \alpha }\]
B) \[-\frac{2}{\sin \alpha }\]
C) \[\frac{1}{\sin \alpha }\]
D) \[-\frac{1}{\sin \alpha }\]
Correct Answer: B
Solution :
\[\sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }}+\sqrt{\frac{1+\cos \alpha }{1-\cos \alpha }}=\frac{1-\cos \alpha +1+\cos \alpha }{\sqrt{1-{{\cos }^{2}}\alpha }}\] \[=\frac{2}{\pm \sin \alpha }\]\[=\frac{2}{-\sin \alpha },\,\,\left( \text{since }\pi <\alpha <\frac{\text{3}\pi }{\text{2}} \right).\]
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