1. The problem is to verify the identity $$\frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta$$. 2. Recall the double-angle formula for sine: $$\sin 2\theta = 2\sin\theta\cos\theta$$. 3. Also recall that $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$. 4. Substitute $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$ into the left side: $$\frac{2\tan\theta}{1+\tan^2\theta} = \frac{2\frac{\sin\theta}{\cos\theta}}{1 + \left(\frac{\sin\theta}{\cos\theta}\right)^2}$$. 5. Simplify the denominator: $$1 + \frac{\sin^2\theta}{\cos^2\theta} = \frac{\cos^2\theta + \sin^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}$$ because $$\sin^2\theta + \cos^2\theta = 1$$. 6. So the expression becomes: $$\frac{2\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos^2\theta}} = 2\frac{\sin\theta}{\cos\theta} \times \cos^2\theta = 2\sin\theta\cos\theta$$. 7. This matches the right side $$\sin 2\theta$$. Therefore, the identity is verified. Final answer: $$\frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta$$.