Proving $\frac {\sin x}{1-\sin x}-\frac {\sin x}{1+\sin x}\equiv 2\tan^2 x$ I need assistance with proving the following identity: 

$$\frac {\sin x}{1-\sin x}-\frac{\sin x}{1+\sin x} \equiv 2\tan^2 x$$

What I have done so far is expanded them:
$$\frac {\sin x\;(1+\sin x)}{(1-\sin x)(1+\sin x)}-\frac {\sin x\;(1-\sin x)}{(1+\sin x)(1-\sin x)}$$
So therefore: 
$$\frac {\sin x+\sin^2 x}{1-\sin^2x}-\frac {\sin x-\sin^2 x}{1-\sin^2 x}$$ 
I'm completely stuck on what to do next.  Any pointers would be appreciated.  Thanks for your time!

As $\cos^2x=(1-\sin x)(1+\sin x)$
$$\dfrac{\sin x}{1-\sin x}=\dfrac{\sin x-1+1}{1-\sin x}=-1+\dfrac{1+\sin x}{\cos^2x}=\tan^2x+\sec x\tan x$$
Similarly, $$\dfrac{\sin x}{1+\sin x}=\sec x\tan x-\tan^2x$$
