$\tan(\pi/2)$ is undefined but $\cot(\pi/2)$ is defined

Question

Given that $$\cot(x) = \dfrac{\cos(x)}{\sin(x)} = \dfrac{1}{\tan(x)},$$ how is it possible for $\cot(x)$ to be defined at points where $\tan(x)$ is undefined? For example, $\tan(\pi/2)$ is undefined yet $\cot(\pi/2) = 0.$

Answer 1

The identity $\cot(x) = \frac{1}{\tan(x)}$ only works where they are both defined. Really, this isn't some sort of definition, but rather a consequence of the fact that $$ \tan(x) = \frac{\sin(x)}{\cos(x)}, \quad \cot(x) = \frac{\cos(x)}{\sin(x)} $$ and if you treat these functions as variables, then it is clear that the identity holds. But this isn't completely accurate, and as you have noted, it doesn't hold for $x$ being an integer multiple of $\frac{\pi}{2}$. In the field of algebraic geometry an identity like $\cot(x) = \frac{1}{\tan(x)}$ would be called a birational equivalence, rather than equality, to reflect this fact. It means the functions are equal wherever they are defined (and that they are defined almost everywhere).

If there is some sort of intuition to get out of this, it could be that $\cot(x)$ and $\frac{1}{\tan(x)}$ can be considered mostly equal as functions, but not necessarily when as evaluations at specific points. So as long as you are manipulating expressions where the functions act like variables, you can consider them equal, but add the constraint $x \neq \frac{n\pi}{2}$ as you go along to help reminding yourself that it is not always true.

Answer 2

Given that $$\cot(x) = \dfrac{\cos(x)}{\sin(x)} = \dfrac{1}{\tan(x)},$$ how is it possible for $\cot(x)$ to be defined at points where $\tan(x)$ is undefined?

1. The usual definition of cotangent is $\cot(x) := \dfrac{\cos(x)}{\sin(x)}. \tag*{}$ In a definition like this, the two sides of the equality have the same domain.

2. The identity $\cot x\equiv\dfrac1{\tan x}\tag*{}$ asserts that its two sides are equal on the intersection of their domains. The two sides of this equality have different domains.

3. Alternatively, to define cotangent in terms of tangent, we can write $\cot x:= \begin{cases} \dfrac{1}{\tan x} & \text{if } x \not\equiv k\frac{\pi}{2} \pmod{\pi}, \\ 0 & \text{if } x \equiv (2k+1)\frac{\pi}{2} \pmod{\pi}. \end{cases}\tag*{}$

Answer 3

Why we say $\tan(\pi/2)$ is undefined. Since $\frac{\sin(\pi/2)}{\cos(\pi/2)}$ and $\cos(\pi/2)=0$, we should say $\tan(\pi/2)$ is undefined. Note that $\frac{*}{0}$ always is undefined.