Trigonometric-Identities Question 387
Question: $ \frac{\sin (B+A)+\cos (B-A)}{\sin (B-A)+\cos (B+A)}= $
[Roorkee 1970; IIT 1966]
Options:
A) $ \frac{\cos B+\sin B}{\cos B-\sin B} $
B) $ \frac{\cos A+\sin A}{\cos A-\sin A} $
C) $ \frac{\cos A-\sin A}{\cos A+\sin A} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{\sin ,(B+A)+\cos ,(B-A)}{\sin ,(B-A)+\cos ,(B+A)} $ $ =\frac{\sin ,(B+A)+\sin ,(90^{o}-\overline{B-A})}{\sin ,(B-A)+\sin ,(90^{o}-\overline{A+B})} $ $ =,\frac{2,\sin ,(A+45^{o}),\cos ,(45^{o}-B)}{2,\sin ,(45^{o}-A),\cos ,(45^{o}-B)} $ $ =\frac{\sin ,(A+45^{o})}{\sin ,(45^{o}-A)}=\frac{\cos A+\sin A}{\cos A-\sin A} $ .
BETA
