Answer
$$\tan(x-y)-\tan(y-x)=\frac{2(\tan x-\tan y)}{1+\tan x\tan y}$$
The equation is an identity.
Work Step by Step
$$\tan(x-y)-\tan(y-x)=\frac{2(\tan x-\tan y)}{1+\tan x\tan y}$$
We expand the left side according to the difference identity for tangents, which states
$$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$
$$X=\tan(x-y)-\tan(y-x)$$
$$X=\frac{\tan x-\tan y}{1+\tan x\tan y}-\frac{\tan y-\tan x}{1+\tan y\tan x}$$
$$X=\frac{\tan x-\tan y}{1+\tan x\tan y}-\Big[\frac{-(\tan x-\tan y)}{1+\tan x\tan y}\Big]$$
$$X=\frac{\tan x-\tan y}{1+\tan x\tan y}+\frac{\tan x-\tan y}{1+\tan x\tan y}$$
$$X=\frac{2(\tan x-\tan y)}{1+\tan x\tan y}$$
So the left side is equal to the right one. This is definitely an identity.
