# $\cot x=0$ is $x$ undefined?

I'm having trouble with finding the values of $x$ when $\cot x=0$

$$\cot x=\frac{1}{\tan x}=0$$ $$\tan x = \frac{1}{0}$$ Which is not possible.

But $$\cot x=\frac{\cos x}{\sin x}=0$$ $$\cos x = 0$$ $$x = 90^\circ$$

So is the value of $x$ undefined or not?

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Probably a good answer would be $$\lim_{x\to\pi/2} \cot{x} = 0$$ – Inquest Jan 21 '13 at 17:39
By this same logic, $\sin(0)$ would not be defined, because $\sin(0)=\dfrac{1}{\csc(0)}$ and $\csc(0)=\dfrac{1}{0}$ is undefined. But $\sin(0)$, like $\cot(90^\circ)$, is perfectly well defined (and equal to $0$). – Jonas Meyer Jan 21 '13 at 17:43
With these trig functions (and, indeed, all analytic functions outside of beginning calculus texts) it is understood that removable singularities should be considered already removed. – GEdgar Jan 21 '13 at 18:07

## 2 Answers

$$\cot x=0\implies \cos x=0\cdot\sin x=0\text{ as } |\sin x|\le 1$$

So, $x=(2n+1)\frac\pi2$ where $n$ is any integer.

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Notice that the points where the tangent function is undefined are exactly the same as the points to which your second method leads you, where the cosine is $0$. In that sense, you're getting the same answer both ways.

I don't often see this mentioned except when I mention it: in trigonometry, as when working with rational functions, it makes sense to take the codomain of the functions to be $\mathbb R\cup\{\infty\}$, where $\infty$ is neither $+\infty$ nor $-\infty$, but rather is a single $\infty$ at both ends of the real line. That makes trigonometric functions everywhere continuous. Then you'd say $1/0=\infty$ and $\tan(\pi/2)=\infty$ and $\cot(\pi/2)=0$.

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I know they are the same points, that is my point (, that they don't provide the same value. – Jonathan. Jan 21 '13 at 18:21
Since tangent and cotangent are each other's reciprocals, and $0$ and $\infty$ are each other's reciprocals, you expect the points where the cotangent is $0$ to be the same as the points where the tangent is $\infty$, and so they are. In that sense, they do "provide the same value[s]". – Michael Hardy Jan 21 '13 at 18:23
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