1. **Problem:** Prove the identity $$2 \sin \theta \cos \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$$. 2. **Recall the formulas:** - Double angle formula for sine: $$\sin 2\theta = 2 \sin \theta \cos \theta$$. - Definition of tangent: $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$. - Pythagorean identity: $$1 + \tan^2 \theta = \sec^2 \theta = \frac{1}{\cos^2 \theta}$$. 3. **Rewrite the right side:** $$\frac{2 \tan \theta}{1 + \tan^2 \theta} = \frac{2 \frac{\sin \theta}{\cos \theta}}{1 + \left(\frac{\sin \theta}{\cos \theta}\right)^2} = \frac{2 \frac{\sin \theta}{\cos \theta}}{1 + \frac{\sin^2 \theta}{\cos^2 \theta}}$$ 4. **Simplify the denominator:** $$1 + \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}$$ 5. **Substitute back:** $$\frac{2 \frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos^2 \theta}} = 2 \frac{\sin \theta}{\cos \theta} \times \cos^2 \theta = 2 \sin \theta \cos \theta$$ 6. **Conclusion:** Both sides are equal, so the identity is proven. $$\boxed{2 \sin \theta \cos \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}}$$