<s>[INST] Solve identity: $\frac {\sin x}{1-\cos x} = \frac {1+\cos x}{\sin x}$ $$\frac {\sin x}{1-\cos x} = \frac {1+\cos x}{\sin x}$$
The only way I can see of doing this is by cross multiplying but isn't that not allowed when trying to prove something?
 [/INST] It's $$\sin^2x=1-\cos^2x$$ or
$$\sin^2x+\cos^2x=1.$$
Also, we have:
$$\frac {\sin x}{1-\cos x}=\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\sin^2\frac{x}{2}}=\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}=\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\frac{1+\cos{x}}{\sin{x}}$$ 
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