Answer
$\frac{\tan x-\tan y}{1-\tan x\tan y}=\frac{\sin(x-y)}{\cos(x+y)}$
Work Step by Step
Start from the left side:
$\frac{\tan x-\tan y}{1-\tan x\tan y}$
Express tangent as sine divided by cosine:
$=\frac{\frac{\sin x}{\cos x}-\frac{\sin y}{\cos y}}{1-\frac{\sin x}{\cos x}*\frac{\sin y}{\cos y}}$
Multiply top and bottom by $\cos x\cos y$:
$=\frac{(\frac{\sin x}{\cos x}-\frac{\sin y}{\cos y})\cos x\cos y}{(1-\frac{\sin x}{\cos x}*\frac{\sin y}{\cos y})\cos x\cos y}$
$=\frac{\sin x\cos y-\cos x\sin y}{\cos x\cos y-\sin x\sin y}$
Use the identities $\sin(x-y)=\sin x\cos y-\cos x\sin y$ and $\cos(x+y)=\cos x\cos y-\sin x\sin y$:
$=\frac{\sin(x-y)}{\cos(x+y)}$
Since this equals the right side, the identity has been proven.
