<s>[INST] 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!
 [/INST] 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 $1-\sin^2x=\cos^2x$.
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