A) \[\frac{b}{a}\]
B) \[\frac{a}{b}\]
C) \[ab\]
D) \[1\]
Correct Answer: B
Solution :
\[\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}\] Applying componendo and dividendo, we get \[\frac{\sin (x+y)+\sin (x-y)}{\sin (x+y)-\sin (x-y)}=\frac{(a+b)+(a-b)}{(a+b)-(a-b)}\] \[\Rightarrow \,\,\frac{2\sin \,x.cosy}{2\cos x.\sin y}=\frac{2a}{2b}\Rightarrow \,\,\tan x.\cot y=\frac{a}{b}\] \[\therefore \,\,\,\frac{\tan x}{\tan y}=\frac{a}{b}\]
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