## Trigonometry (11th Edition) Clone

$$\frac{\tan(x+y)-\tan y}{1+\tan(x+y)\tan y}=\tan x$$ After proving that left side is equal to right side as below, we can conclude that the equation is an identity.
$$\frac{\tan(x+y)-\tan y}{1+\tan(x+y)\tan y}=\tan x$$ From the left side: $$X=\frac{\tan(x+y)-\tan y}{1+\tan(x+y)\tan y}$$ - Recall the difference identity for tangent: $$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$ For the whole $X$, we can apply the identity with $A=x+y$ and $B=y$, leading to $$X=\tan[(x+y)-y]$$ $$X=\tan(x+y-y)$$ $$X=\tan x$$ That concludes $$\frac{\tan(x+y)-\tan y}{1+\tan(x+y)\tan y}=\tan x$$ meaning the equation is an identity.