Correct Answer: 1

Solution : Given:
$\frac{\sin (x-y)}{\sin (x+y)}(\frac{\tan x+\tan y}{\tan x–\tan y})$
Using $\sin(x+y)=\sin x\cos y+\cos x\sin y$ and $\sin(x-y)=\sin x\cos y-\cos x\sin y$, we get:
$=(\frac{\sin x\cos y–\cos x\sin y}{\sin x\cos y+\cos x\sin y})(\frac{\tan x+\tan y}{\tan x–\tan y})$
Dividing both numerator and denominator by $\cos x\cos y$, we get,
$=(\frac{\frac{\sin x\cos y–\cos x\sin y}{\cos x\cos y}}{\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y}})(\frac{\tan x+\tan y}{\tan x–\tan y})$
$= (\frac{\tan x–\tan y}{\tan x+\tan y})(\frac{\tan x+\tan y}{\tan x–\tan y})$
$= 1$
Hence, the correct answer is 1.