Prove this identity... $$\frac{\sin 2x}{1+\cos 2x} \times \frac{\cos x}{1+\cos x}=\tan\frac{x}{2}$$
This is what I've done:
$$\frac{2\sin x \cos x}{1+\cos^2 x-\sin^2 x} \times \frac{\cos x}{1+\cos x}=$$
$$\frac{2\sin x \cos x}{2\cos^2 x} \times \frac{\cos x}{1+\cos x}=$$
$$\frac{\sin x}{1+\cos x}$$
I have no idea what to do next.
edit:
Solution:
$$\frac{\sin2\frac{x}{2}}{1+\cos2\frac{x}{2}}=$$
$$\frac{2\sin\frac{x}{2}cos\frac{x}{2}}{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}=$$
$$\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}=$$
$$\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\tan\frac{x}{2}$$

$\cos 2\theta = 2\cos^2 \theta - 1 \Rightarrow 1 + \cos 2\theta = 2\cos^2 \theta$. Therefore:
$$\dfrac{\sin 2x}{1 + \cos 2x} \times \dfrac{\cos x}{1 + \cos x}\\
= \dfrac{\not 2\sin x \not\cos x}{\not 2 \not\cos^2 x} \times \dfrac{\not\cos x}{2 \cos^2 (x/2)}\\
= \dfrac{\not 2\sin(x/2)\not\cos(x/2)}{\not 2\cos^{\not 2} (x/2)}\\
= \boxed{\tan \dfrac{x}{2}}
$$
Note: In the above, the $\not\square$s denote cancellations.
