1. **State the problem:** Prove the identity $$\frac{\sin 2x}{1 + \cos 2x} \cdot \frac{\cos x}{1 + \cos x} = \tan \frac{x}{2}$$. 2. **Recall key formulas:** - Double angle formulas: $$\sin 2x = 2 \sin x \cos x$$ and $$\cos 2x = 2 \cos^2 x - 1$$. - Half-angle formula for tangent: $$\tan \frac{x}{2} = \frac{\sin x}{1 + \cos x}$$. 3. **Simplify the left side:** Start with $$\frac{\sin 2x}{1 + \cos 2x} \cdot \frac{\cos x}{1 + \cos x}$$. Substitute double angle formulas: $$= \frac{2 \sin x \cos x}{1 + (2 \cos^2 x - 1)} \cdot \frac{\cos x}{1 + \cos x}$$ Simplify denominator: $$1 + 2 \cos^2 x - 1 = 2 \cos^2 x$$ So expression becomes: $$= \frac{2 \sin x \cos x}{2 \cos^2 x} \cdot \frac{\cos x}{1 + \cos x}$$ Simplify the fraction: $$= \frac{\sin x}{\cos x} \cdot \frac{\cos x}{1 + \cos x}$$ Cancel $$\cos x$$: $$= \frac{\sin x}{1 + \cos x}$$ 4. **Recognize the half-angle formula:** $$\frac{\sin x}{1 + \cos x} = \tan \frac{x}{2}$$. 5. **Conclusion:** The left side simplifies exactly to the right side, proving the identity. **Final answer:** $$\frac{\sin 2x}{1 + \cos 2x} \cdot \frac{\cos x}{1 + \cos x} = \tan \frac{x}{2}$$.