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!
 A: 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.
A: Your idea is good and you were close but then I don't understand what you did:
$$\frac{\sin x}{1-\sin x}-\frac{\sin x}{1+\sin x}=\frac{\sin x+\sin^2x-\sin x+\sin^2x}{1-\sin^2x}=$$
$$=2\frac{\sin^2x}{\cos^2x}=2\tan^2x$$
A: You have
$$\frac{\sin x}{1-\sin x}-\frac{\sin x}{1+\sin x}=\frac {\sin x+\sin^2x}{1-\sin^2x} -\frac {\sin x-\sin^2x}{1-\sin^2x}$$
Then, use that $\frac{B}{A}-\frac CA=\frac{B-C}{A}$ to get $$\frac {\sin x+\sin^2x-(\sin x-\sin^2x)}{1-\sin^2x}=\frac{2\sin^2x}{1-\sin^2x}$$
Now use $1-\sin^2x=\cos^2x$.
A: 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, $$\dfrac{\sin x}{1+\sin x}=\sec x\tan x-\tan^2x$$
