Answer
$$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$$
The proof is below in the work step by step.
Work Step by Step
$$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$$
We examine the right side first.
$$A=\frac{2\tan\theta}{1+\tan^2\theta}$$
As from quotient identity: $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$
so, $$A=\frac{\frac{2\sin\theta}{\cos\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}$$
$$A=\frac{\frac{2\sin\theta}{\cos\theta}}{\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}}$$
Now recall that $\cos^2\theta+\sin^2\theta=1$. Thus,
$$A=\frac{\frac{2\sin\theta}{\cos\theta}}{\frac{1}{\cos^2\theta}}$$
$$A=\frac{2\sin\theta\cos^2\theta}{\cos\theta\times1}$$
$$A=2\sin\theta\cos\theta$$
Finally, we can rewrite $2\sin\theta\cos\theta$ into $\sin2\theta$, according to double-angle identity for sines.
$$A=\sin2\theta$$
Therefore, $$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$$
The equation is an identity.
