A) \[\pm \sqrt{\frac{1-\sin A}{1+\sin A}}\]
B) \[\pm \sqrt{\frac{1+\sin A}{1-\sin A}}\]
C) \[\pm \sqrt{\frac{1-\cos A}{1+\cos A}}\]
D) \[\pm \sqrt{\frac{1+\cos A}{1-\cos A}}\]
Correct Answer: C
Solution :
\[\tan \left( \frac{A}{2} \right)=\frac{\sin (A/2)}{\cos (A/2)}=\pm \sqrt{\frac{(1-\cos A)/2}{(1+\cos A)/2}}=\pm \sqrt{\frac{1-\cos A}{1+\cos A}}\].
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