Answer
$$\frac{\tan(x+y)-\tan y}{1+\tan(x+y)\tan y}=\tan x$$
After proving that left side is equal to right side as below, we can conclude that the equation is an identity.
Work Step by Step
$$\frac{\tan(x+y)-\tan y}{1+\tan(x+y)\tan y}=\tan x$$
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.
