Answer
$\frac{\sin(x-y)}{\sin(x+y)}=\frac{\tan x-\tan y}{\tan x+\tan y}$
Work Step by Step
Start with the right side:
$\frac{\tan x-\tan y}{\tan x+\tan y}$
Express it in terms of sine and cosine:
$=\frac{\frac{\sin x}{\cos x}-\frac{\sin y}{\cos y}}{\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}}$
Multiply top and bottom by $\cos x\cos y$:
$=\frac{\frac{\sin x}{\cos x}-\frac{\sin y}{\cos y}}{\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}}*\frac{\cos x\cos y}{\cos x\cos y}$
$=\frac{\sin x\cos y-\cos x\sin y}{\sin x\cos y+\cos x\sin y}$
Use sum and difference identities for sine to simplify:
$=\frac{\sin(x-y)}{\sin(x+y)}$
Since this equals the left side, the identity has been proven.
