Dividing by possibly zero value when proving an identity? In my trigonometry course, we're currently proving identities. I'm wonder if I can divide by something that could be zero while proving it. For example, $\sin{x}$, is it still a valid proof if I divide by it?
edit: Sorry, what I really meant to ask was if it's alright to divide by a variable when DISproving an identity. I would assume it's alright if the disproof stands, right?

@Hurkyl is right about dividing by zero in the common case, but I'd clarify something for your specific case.
You've said that you're dealing with identities, not equations (the identity is like $\sin{2x} = 2\sin{x}\cos{x}$). In that sense, dividing by something which could be zero (but only at some points: such as $\sin{x}$, but not $\min(0, |x|-1)$) is legitimate, because:


*

*On the both sides of the equation you have functions with no discontinuities (except for the countable set of points where the limits from both sides are equal to $\infty$, and the value in the point itself is not defined);

*When dividing by e.g. $\sin{x}$, you achieve the continuous (in the sense above) function;

*Given two continuous (in the sense above) functions which are equal almost everywhere (e.g. everywhere except for the countable set of points), one can deduce by continuity that the functions are equal everywhere where they're defined, so it is safe to say two functions are equal / identical.
Thus you can prove e.g. that $\tan^2{x} = \dfrac{\sin^2{x}}{\cos^2{x}} = \dfrac{1-\cos^2{x}}{\cos^2{x}} = \dfrac{1}{\cos^2{x}}-1$ without having to bother about the points where $\cos{x}$ is zero.
