1. The problem is to prove the identity $$\frac{\tan 4A - \tan 3A}{1 + \tan 4A \tan 3A} = \tan A.$$\n\n2. Recall the tangent subtraction formula: $$\tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}.$$\nThis formula states that the tangent of the difference of two angles equals the fraction on the left side of the given equation.\n\n3. By comparing, we see that $$x = 4A$$ and $$y = 3A.$$\nTherefore, $$\frac{\tan 4A - \tan 3A}{1 + \tan 4A \tan 3A} = \tan(4A - 3A) = \tan A.$$\n\n4. Hence, the identity is proven by direct application of the tangent subtraction formula.