A) \[\frac{\cos A}{1+\sin A}\]
B) \[\frac{\sin A}{1-\cos A}\]
C) \[\frac{\tan A}{1+\tan A}\]
D) \[\frac{\tan A}{1+\cos A}\]
Correct Answer: A
Solution :
\[\frac{1-\sin A}{\cos A}=\frac{(1-\sin A)\,(1+\sin A)}{\cos A(1+\sin A)}\] \[=\frac{1-{{\sin }^{2}}A}{\cos A(1+\sin A)}\] \[=\frac{{{\cos }^{2}}A}{\cos A(1+\sin A)}\] \[=\frac{\cos A}{1+\sin A}\]
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