Domain of $\frac {\cos x}{\sin x}$ vs $\frac {1}{\tan x}$? I was helping a precalculus student  graph a function involving $\cot x$, and I realized that these two different definitions (which I thought equivalent) are slightly different.
When finding the domain of  $\dfrac {\cos x}{\sin x}$, we only exclude the zeroes of $\sin x$. When finding the domain of $\dfrac {1}{\tan x} = \dfrac {1}{\frac {\sin x}{\cos x}}$, we must concern ourselves with the zeroes of both $\sin x$ and $\cos x$. My friend said the definitions are equivalent, because $\dfrac {\cos x}{\sin x} = \dfrac {1}{\frac {\sin x}{\cos x}}$ even when $\cos x = 0$, but I don't believe that's true. Which one of us is right?
 A: You can write x as $\frac{1}{\frac{1}{x}}$ and it poses the same problem. When written in the latter form there is a removable singularity at x=0. We call it "removable" since the function y=x would be continious at x=0 if we replaced the missing point with the $\lim_{x\rightarrow 0}x$. The same applies for the multiple singularities generated when we write $\frac{1}{\tan x}$ as $\frac{1}{\frac{\sin x}{\cos x}}$
A: You are correct, and so is your reasoning. If $\cos(x)=0$, then $$\frac{\cos(x)}{\sin(x)}$$ is zero divided by a nonzero number, which is defined under real arithmetic, and is equal to zero. On the other hand, $$\frac{1}{\frac{\sin(x)}{\cos(x)}}$$ is one divided by $$\frac{\sin(x)}{\cos(x)}.$$ The latter is a nonzero number divided by zero. This is not defined. Therefore the value of $$\frac{1}{\frac{\sin(x)}{\cos(x)}}$$ is similarly undefined.
A: You are both right.
There are two different useful conventions, which could roughly be described as:


*

*Exactly propagate the domain of definition of an expression in precisely the way you describe.

*Normalize the result of division, cleaning up the result of an expression by doing things like continuously extending through removable discontinuities, and maybe to $\pm \infty$ as well.


You are right when interpreting the formula via the first convention, and your friend is right when interpreting the formula via the second convention
A: Domain of $\frac {\cos (x)}{\sin (x)} $
$$=\{x\in\mathbb R \,:\;\sin (x)\ne 0\}$$
while
Domain of $\frac {1}{\tan (x)} $
$$=\{x\in \mathbb R \;:\: \sin (x)\cos (x)\ne 0\}$$
