#### Answer

$\tan(x-y)-\tan(y-x)=\frac{2(\tan x-\tan y)}{1+\tan x\tan y}$

#### Work Step by Step

Start with the left side:
$\tan(x-y)-\tan(y-x)$
$=\tan(x-y)-\tan(-(x-y))$
Use the fact that tangent is an odd function, i.e. $\tan(-\theta)=-\tan\theta$ for all $\theta$:
$=\tan(x-y)-(-\tan(x-y))$
$=2\tan(x-y)$
Use the difference formula for tangent on page 673:
$=2*\frac{\tan x-\tan y}{1+\tan x\tan y}$
$=\frac{2(\tan x-\tan y)}{1+\tan x\tan y}$
Since this equals the right side, the identity has been proven.