Fundamental Period of $\tan x \cot x$

Question

What is the period of $\tan x \cot x?$ I was given this question today. What I did was simplify the expression , and it reduced to a constant function. So it had no fundamental period.

But my teacher told me that the answer was $\frac{\pi}{2}$. How is it so?

Answer 1

$$f(x) = \dfrac{\tan x}{\tan x}\neq 1$$ $$f(x) = 1, x\neq k\dfrac{\pi}{2}, k\in\mathbb{I}\\$$

The period is $\pi / 2$, since the function isn't defined for integral multiple of pi/2. So as egreg also mentions, it will have holes in graph.

For a function to be periodic, it must have translational symmetry in its graph.

Answer 2

This is $1$, so it is constant.

Are you sure that you have the right problem?

Might the problem be $\tan(\cot(x))$?

Answer 3

A function $f : A \to B$ is periodic with period $p$ if, for all $x \in A$, $$f(x + p) = f(x)$$ and $p$ is the least such value for which this is true. For $$f(x) = \tan x \cot x,$$ we see that $f$ is well-defined on the domain $$A = \{x \in \mathbb R : 2x/\pi \not\in \mathbb Z\},$$ that is to say, all reals which are not integer multiples of $\pi/2$. On this domain, we can state $f(x) = 1$: another way to put it is, $$f(x) = \begin{cases} 1, & x \in A \\ \text{undefined}, & x \not\in A. \end{cases}$$ Now suppose that $0 < p < \pi/2$. Then we would have $f(x+p) = 1 = f(x)$ whenever $x \in A$ and $x+p \in A$. But if $x+p \not\in A$ while $x \in A$, we would have a violation of the periodicity condition. Yet a situation arises for $x = \pi/2 - p$: since by construction $0 < p < \pi/2$, $x = \pi/2 - p \in A$, yet $x + p = \pi/2 \not \in A$. Consequently, $f$ cannot be periodic with $p < \pi/2$.

It is rather straightforward to continue to reason that $f$ is periodic with $p = \pi/2$. For example, suppose $x \not\in A$; then $2x/\pi \in \mathbb Z$. Then $$\frac{2(x+\pi/2)}{\pi} = \frac{2x}{\pi} + 1 \in \mathbb Z,$$ consequently $x+\pi/2 \not\in A$. And since each step is reversible, we conclude $x + \pi/2 \not\in A$ implies $x \not \in A$; hence $x \in A$ if and only if $x + \pi/2 \in A$. Then apply the period condition to conclude $f(x+\pi/2) = 1 = f(x)$ for all $x \in A$.