Trigonometric Identities Question 249
Question: If $ \frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}, $ then what is $ \frac{\tan x}{\tan y} $ equal to?
Options:
A) $ \frac{b}{a} $
B) $ \frac{a}{b} $
C) $ ab $
D) $ 1 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b} $ Applying componendo and dividendo, we get $ \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.cosy}{2\cos x.\sin y}=\frac{2a}{2b}\Rightarrow \tan x.\cot y=\frac{a}{b} $
$ \therefore ,\frac{\tan x}{\tan y}=\frac{a}{b} $
BETA
