1. **State the problem:** Prove the trigonometric identity $$\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$$. 2. **Recall the formula for tangent of a sum:** $$\tan(x+y) = \frac{\sin(x+y)}{\cos(x+y)}$$. 3. **Use the sine and cosine addition formulas:** $$\sin(x+y) = \sin x \cos y + \cos x \sin y$$ $$\cos(x+y) = \cos x \cos y - \sin x \sin y$$ 4. **Substitute these into the tangent expression:** $$\tan(x+y) = \frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y - \sin x \sin y}$$ 5. **Divide numerator and denominator by $\cos x \cos y$ to express in terms of tangent:** $$\tan(x+y) = \frac{\frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}}{1 - \frac{\sin x}{\cos x} \cdot \frac{\sin y}{\cos y}} = \frac{\tan x + \tan y}{1 - \tan x \tan y}$$ 6. **Conclusion:** We have shown that $$\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$$, which completes the proof.