<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] Set $t=\tan \frac x2$ so that $\sin x = \cfrac {2t}{1+t^2}$ and $\cos x = \cfrac {1-t^2}{1+t^2}$ (Weierstrass substitution) or $$(1+t^2)\sin x =2t; (1+t^2)\cos x=1-t^2$$
Since $1+t^2\ge 1 \gt 0$ we can multiply the original fractions by $1=\cfrac {1+t^2}{1+t^2}$ to obtain $$\frac {\sin x}{1-\cos x}=\frac {2t}{2t^2}=\frac 2{2t}=\frac {1+t^2+1-t^2}{2t}=\frac {1+\cos x}{\sin x}$$
The only issue is cancelling a factor of $\frac tt$ which is undefined when $t=0$, but then one of the original fractions reduces to $\frac 00$.
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