A) \[\sec A-\tan A\]
B) \[\text{cosec}\,A+\cot A\]
C) \[\tan \left( \frac{\pi }{4}-\frac{A}{2} \right)\]
D) \[\tan \left( \frac{\pi }{4}+\frac{A}{2} \right)\]
Correct Answer: D
Solution :
\[\frac{\cos A}{1-\sin A}=\frac{\cos A(1+\sin A)}{{{\cos }^{2}}A}=\frac{(1+\sin A)}{\cos A}\] \[=\frac{{{\left( \cos \frac{A}{2}+\sin \frac{A}{2} \right)}^{2}}}{\left( \cos \frac{A}{2}+\sin \frac{A}{2} \right)\,\left( \cos \frac{A}{2}-\sin \frac{A}{2} \right)}=\frac{\cos \frac{A}{2}+\sin \frac{A}{2}}{\cos \frac{A}{2}-\sin \frac{A}{2}}\] \[=\frac{1+\tan \frac{A}{2}}{1-\tan \frac{A}{2}}\], \[\left( \text{Dividing}\,{{N}^{r}}\,\text{and}\,{{D}^{r}}\,\text{by}\,\cos \frac{A}{2} \right)\] \[=\tan \left( \frac{\pi }{4}+\frac{A}{2} \right)\].
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