#### Answer

$\frac{\sin(x+y)}{\cos(x-y)}=\frac{\cot x+\cot y}{1+\cot x\cot y}$

#### Work Step by Step

Start with the right side:
$\frac{\cot x+\cot y}{1+\cot x\cot y}$
Express it in terms of sine and cosine:
$=\frac{\frac{\cos x}{\sin x}+\frac{\cos y}{\sin y}}{1+\frac{\cos x}{\sin x}*\frac{\cos y}{\sin y}}$
Multiply top and bottom by $\sin x\sin y$:
$=\frac{\frac{\cos x}{\sin x}+\frac{\cos y}{\sin y}}{1+\frac{\cos x}{\sin x}*\frac{\cos y}{\sin y}}*\frac{\sin x\sin y}{\sin x\sin y}$
$=\frac{\cos x\sin y+\sin x\cos y}{\sin x\sin y+\cos x\cos y}$
$=\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y+\sin x\sin y}$
Use sum and difference identities for sine and cosine to simplify:
$=\frac{\sin(x+y)}{\cos(x-y)}$
Since this equals the left side, the identity has been proven.