Trigonometric Identities Question 235
Question: $ \frac{\cos A}{1-\sin A}= $
Options:
A) $ \sec A-\tan A $
B) $ cosec,A+\cot A $
C) $ \tan ( \frac{\pi }{4}-\frac{A}{2} ) $
D) $ \tan ( \frac{\pi }{4}+\frac{A}{2} ) $
Show Answer
Answer:
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{{{( \cos \frac{A}{2}+\sin \frac{A}{2} )}^{2}}}{( \cos \frac{A}{2}+\sin \frac{A}{2} ),( \cos \frac{A}{2}-\sin \frac{A}{2} )}=\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}} $ , $ ( Dividing,N^{r},and,D^{r},by,\cos \frac{A}{2} ) $ $ =\tan ( \frac{\pi }{4}+\frac{A}{2} ) $ .
BETA
