## Trigonometry (11th Edition) Clone

$$\tan(x-y)-\tan(y-x)=\frac{2(\tan x-\tan y)}{1+\tan x\tan y}$$ The equation is an identity.
$$\tan(x-y)-\tan(y-x)=\frac{2(\tan x-\tan y)}{1+\tan x\tan y}$$ 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.