## Precalculus (6th Edition)

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