Trigonometric-Identities Question 388
Question: If $ \frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}, $ then $ \frac{\tan x}{\tan y} $ is equal to
Options:
A) $ \frac{b}{a} $
B) $ \frac{a}{b} $
C) $ ab $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{\sin ,(x+y)}{\sin ,(x-y)}=\frac{a+b}{a-b} $
$ \Rightarrow \frac{\sin ,(x+y)+\sin ,(x-y)}{\sin ,(x+y)-\sin ,(x-y)}=\frac{(a+b)+(a-b)}{(a+b)-(a-b)} $
$ \Rightarrow \frac{2,\sin x,\cos y}{2,\cos x,\sin y}=\frac{2a}{2b},\Rightarrow \frac{\tan x}{\tan y}=\frac{a}{b} $ .
BETA
