What is the domain and range of the function $f(x)= \frac{1}{\tan{x}}$? The function:
$f(x)= \frac{1}{\tan{x}}$
We've written in class that the domain of this function is $ \{x|x \ne n\pi\} $ where $n$ is an integer, and that the range is equal to $R - \{0\}$
Now what I dont underatand is that if the range doesn't include $0$ , shouldn't the domain exclude $ x=\frac{\pi}{2}+ n\pi $, where n is an integer as well?
 A: Yes. The only way you could include $\frac{\pi}{2}+n\pi$ in the domain would be if you are allowing "$\infty$" as a value, and calculating $\dfrac{1}{\infty}=0$, in which case to be consistent, $0$ would also have to be in the range.  Making that choice is done more commonly when working with meromorphic functions in complex analysis.  But in terms of functions with real number inputs and outputs, if $\tan(x)$ isn't defined, then neither is $\dfrac{1}{\tan(x)}$.  And if $a$ is a real number, then $\dfrac{1}{a}\neq 0$, which is the reason $0$ is not in the range in that context.  So the statement of the domain is missing just what you said.

Added: However, the points of the form $\frac{\pi}{2}+n\pi$ are "removable singularities", and this function is "essentially" equal to $\cot(x)$.  If it were simply $\cot(x)$, then it would have the domain you stated, and the range would include $0$.  In some cases removing removable singularities is a good default position to take, but different instructors may disagree.
