Trigonometric Identities Question 339
Question: $ \frac{\tan A+\sec A-1}{\tan A-\sec A+1}= $
Options:
A) $ \frac{1-\sin A}{\cos A} $
B) $ \frac{1-\cos A}{\sin A} $
C) $ \frac{1+\sin A}{\cos A} $
D) $ \frac{1+\cos A}{\sin A} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{\tan A+\sec A-1}{\tan A-\sec A+1} $ $ =\frac{\sin A-\cos A+1}{\sin A-1+\cos A}=\frac{\sin A+(1-\cos A)}{\sin A-(1-\cos A)} $ $ =\frac{2\sin \frac{A}{2}\cos \frac{A}{2}+2{{\sin }^{2}}\frac{A}{2}}{2\sin \frac{A}{2}\cos \frac{A}{2}-2{{\sin }^{2}}\frac{A}{2}} $ $ =\frac{\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} )}^{2}}}{{{\cos }^{2}}\frac{A}{2}-{{\sin }^{2}}\frac{A}{2}} $ $ =\frac{1+\sin A}{\cos A} $ .
BETA
