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?
 A: Note that the identity doesn't work for some values of $x$, specifically when $\sin(x) = 0$ or when $1 - \cos(x) = 0$. However, if $1 - \cos(x) = 0$, then
\begin{align*}
& \cos(x) = 1 \\
\implies& \cos^2(x) = 1 \\
\implies& \sin^2(x) = 1 - \cos^2(x) = 0 \\
\implies& \sin(x) = 0,
\end{align*}
so really the identity only fails if $\sin(x) = 0$.
Otherwise, let's suppose $\sin(x) \neq 0$. By the above argument, then $1 - \cos(x) \neq 0$. We start with the known identity,
$$\sin^2(x) = 1 - \cos^2(x).$$
Then, using difference of two squares,
$$\sin^2(x) = (1 - \cos(x))(1 + \cos(x)).$$
Since $\sin(x) \neq 0$, we may divide both sides by it:
$$\sin(x) = \frac{(1 - \cos(x))(1 + \cos(x))}{\sin(x)}.$$
Since $1 - \cos(x) \neq 0$, we may also divide by it:
$$\frac{\sin(x)}{1 - \cos(x)} = \frac{1 + \cos(x)}{\sin(x)}.$$

Note that I've essentially used cross multiplication, but in reverse. I've started with $\sin^2(x) = 1 - \cos^2(x)$, the result that we obtain by cross multiplying the above expression. I've then carefully undone the process of cross multiplying, to obtain the equality we wanted to prove.
Why do I write it out this way? Because it's technically a fallacy to start by manipulating the expression you want to prove, obtain something known to be true, and therefore conclude the original expression is true. It's an example of the fallacy of affirming the consequent.
A more blatant example is the following: Suppose $-1 = 1$. Then squaring both sides yields $(-1)^2 = 1^2$, which implies $1 = 1$, which is true. Thus, $-1 = 1$ (but not really).
In our example, there are similar, but lesser problems. Note that the identity we wanted to prove didn't work for all $x$, but cross-multiplying gave us an expression that did work for all $x$. It's not clear, when using cross multiplying (when forming the argument backwards) why there were any problems in the original expression. It's only when you start from $\sin^2(x) = 1 - \cos^2(x)$ and start dividing, that you run into issues with potentially dividing by $0$.
That might be why you've got it in your head that you're not allowed to cross-multiply. The thing is, cross-multiplying is a perfectly valid thing to do in a proof, but the more important thing is to always start with something you know to be true (e.g. $\sin^2(x) = 1 - \cos^2(x)$) and work your way to the conclusion that you want.
A: Yes, cross-multiplying is perfectly allowed.
You start with $$\frac {\sin x}{1-\cos x} = \frac {1+\cos x}{\sin x}$$
$$\iff \sin² x = (1- \cos x)(1+ \cos x) = 1 - \cos² x, \sin x \neq 0, \cos x  \neq 1$$ 
and the last expression is always true, such that the expression you started with is always true (in particular, we use the $\Leftarrow$ implication) 
A: It's $$\sin^2x=1-\cos^2x$$ or
$$\sin^2x+\cos^2x=1.$$
Also, we have:
$$\frac {\sin x}{1-\cos x}=\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\sin^2\frac{x}{2}}=\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}=\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\frac{1+\cos{x}}{\sin{x}}$$ 
A: 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 $\frac 00$.
