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!

Hint: Simplify the original expression, 
 You get  $\cos^2 (x)$ in denominator and $2\sin^2 (x)$ in numerator:
By cross multiplying, you get
$$\dfrac{\sin x+\sin^2 x - \sin x + \sin^2 x }{1-\sin^2 x} = \dfrac{2\sin^2 x}{1-\sin^2 x}$$
Now use $$1-\sin^2 x = \cos^2 x$$ to get the answer.
