A) \[\frac{b}{a}\]
B) \[\frac{a}{b}\]
C) \[ab\]
D) None of these
Correct Answer: B
Solution :
\[\frac{\sin \,(x+y)}{\sin \,(x-y)}=\frac{a+b}{a-b}\] \[\Rightarrow \,\,\frac{\sin \,(x+y)+\sin \,(x-y)}{\sin \,(x+y)-\sin \,(x-y)}=\frac{(a+b)+(a-b)}{(a+b)-(a-b)}\] \[\Rightarrow \,\,\frac{2\,\sin x\,\cos y}{2\,\cos x\,\sin y}=\frac{2a}{2b}\,\Rightarrow \,\,\frac{\tan x}{\tan y}=\frac{a}{b}\].
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