# 8.2: The Tangent Function

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

Let

$A=\left\{\frac{\pi}{2}+n \pi: n \in \mathbb{Z}\right\}$

and $$D=\mathbb{R} \backslash A$$. Let

$t:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow \mathbb{R}$

be the inverse of the arctangent function. Note that $$t$$ is increasing and differentiable on $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) .$$ We may extend $$t$$ to a function on $$D$$ as follows: For any $$x \in D,$$ let

$g(x)=\sup \left\{n: n \in \mathbb{Z},-\frac{\pi}{2}+n \pi<x\right\}$

and define $$T(x)=t(x-g(x) \pi)$$.

## Definition

With the notation of the above discussion, for any $$x \in D,$$ we call the value $$T(x)$$ the tangent of $$x,$$ which we denote $$\tan (x) .$$

## Proposition $$\PageIndex{1}$$

The tangent function has domain $$D$$ (as defined above), range $$\mathbb{R},$$ and is differentiable at every point $$x \in D .$$ Moreover, the tangent function is increasing on each interval of the form

$\left(-\frac{\pi}{2}+n \pi, \frac{\pi}{2}+n \pi\right),$

$$n \in \mathbb{Z},$$ with

$\tan \left(\left(\frac{\pi}{2}+n \pi\right)+\right)=-\infty$

and

$\tan \left(\left(\frac{\pi}{2}+n \pi\right)-\right)=+\infty .$

Proof

These results follow immediately from our definitions. $$\quad$$ Q.E.D.

## Definition

Let $$E \subset \mathbb{R}$$. We say a function $$f: E \rightarrow \mathbb{R}$$ is periodic if there exists a real number $$p>0$$ such that, for each $$x \in E, x+p \in E$$ and $$f(x+p)=f(x) .$$ We say $$p$$ is the period of a periodic function $$f$$ if $$p$$ is the smallest positive number for which $$f(x+p)=f(x)$$ for all $$x \in E .$$

## Proposition $$\PageIndex{2}$$

The tangent function has period $$\pi .$$

Proof

The result follows immediately from our definitions. $$\quad$$ Q.E.D.

## Proposition $$\PageIndex{3}$$

For any $$x, y \in D$$ with $$x+y \in D$$,

$\tan (x+y)=\frac{\tan (x)+\tan (y)}{1-\tan (x) \tan (y)}.$

Proof

Suppose $$y_{1}, y_{2} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$ with $$y_{1}+y_{2} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) .$$ Let $$x_{1}=\tan \left(y_{1}\right)$$ and $$x_{2}=\tan \left(y_{2}\right) .$$ Note that if $$x_{1}>0,$$ then $$x_{1} x_{2} \geq 1$$ would imply that

$x_{2} \geq \frac{1}{x_{1}},$

which in turn implies that

\begin{aligned} y_{1}+y_{2} &=\arctan \left(x_{1}\right)+\arctan \left(x_{2}\right) \\ & \geq \arctan \left(x_{1}\right)+\arctan \left(\frac{1}{x_{1}}\right) \\ &=\frac{\pi}{2}, \end{aligned}

contrary to our assumptions. Similarly, if $$x_{1}<0,$$ then $$x_{1} x_{2} \geq 1$$ would imply that

$x_{2} \leq \frac{1}{x_{1}},$

which in turn implies that

\begin{aligned} y_{1}+y_{2} &=\arctan \left(x_{1}\right)+\arctan \left(x_{2}\right) \\ & \leq \arctan \left(x_{1}\right)+\arctan \left(\frac{1}{x_{1}}\right) \\ &=-\frac{\pi}{2}, \end{aligned}

contrary to our assumptions. Thus we must have $$x_{1} x_{2}<1 .$$ Moreover, suppose $$u$$ is a number between $$-x_{1}$$ and $$x_{2} .$$ If $$x_{1}>0,$$ then

$x_{2}<\frac{1}{x_{1}},$

and so

$u<\frac{1}{x_{1}}.$

If $$x_{1}<0,$$ then

$x_{2}>\frac{1}{x_{1}},$

and so

$u>\frac{1}{x_{1}}.$

Now let

$x=\frac{x_{1}+x_{2}}{1-x_{1} x_{2}}.$

We want to show that

$\arctan (x)=\arctan \left(x_{1}\right)+\arctan \left(x_{2}\right),$

which will imply that

$\frac{\tan \left(y_{1}\right)+\tan \left(y_{2}\right)}{1-\tan \left(y_{1}\right) \tan \left(y_{2}\right)}=\tan \left(y_{1}+y_{2}\right).$

We need to compute

$\arctan (x)=\arctan \left(\frac{x_{1}+x_{2}}{1-x_{1} x_{2}}\right)=\int_{0}^{\frac{x_{1}+x_{2}}{1-x_{1} x_{2}}} \frac{1}{1+t^{2}} d t.$

Let

$t=\varphi(u)=\frac{x_{1}+u}{1-x_{1} u},$

where $$u$$ varies between $$-x_{1},$$ where $$t=0,$$ and $$x_{2},$$ where $$t=x .$$ Now

$\varphi^{\prime}(u)=\frac{\left(1-x_{1} u\right)-\left(x_{1}+u\right)\left(-x_{1}\right)}{\left(1-x_{1} u\right)^{2}}=\frac{1+x_{1}^{2}}{\left(1-x_{1} u\right)^{2}},$

which is always positive, thus showing that $$\varphi$$ is an increasing function, and

\begin{aligned} \frac{1}{1+t^{2}} &=\frac{1}{1+\left(\frac{x_{1}+u}{1-x_{1} u}\right)^{2}} \\ &=\frac{\left(1-x_{1} u\right)^{2}}{\left(1-x_{1} u\right)^{2}+\left(x_{1}+u\right)^{2}} \\ &=\frac{\left(1-x_{1} u\right)^{2}}{\left(1+x_{1}^{2}\right)\left(1+u^{2}\right)}. \end{aligned}

Hence

\begin{aligned} \arctan (x) &=\int_{-x_{1}}^{x_{2}} \frac{1}{1+u^{2}} d u \\ &=\int_{-x_{1}}^{0} \frac{1}{1+u^{2}} d u+\int_{0}^{x_{2}} \frac{1}{1+u^{2}} d u \\ &=-\int_{0}^{-x_{1}} \frac{1}{1+u^{2}} d u+\arctan \left(x_{2}\right) \\ &=-\arctan \left(-x_{1}\right)+\arctan \left(x_{2}\right) \\ &=\arctan \left(x_{1}\right)+\arctan \left(x_{2}\right). \end{aligned}

Now suppose $$y_{1}, y_{2} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$ with $$y_{1}+y_{2}>\frac{\pi}{2} .$$ Then $$y_{1}+y_{2} \in\left(\frac{\pi}{2}, \pi\right)$$,$$x_{1}>0, x_{2}>0,$$ and

$x_{2}>\frac{1}{x_{1}}.$

With $$u$$ and $$x$$ as above, note then that as $$u$$ increases from $$-x_{1}$$ to $$\frac{1}{x_{1}}, t$$ increases from 0 to $$+\infty,$$ and as $$u$$ increases from $$\frac{1}{x_{1}}$$ to $$x_{2}, t$$ increases from $$-\infty$$ to $$x .$$

Hence we have

\begin{aligned} \arctan (x)+\pi &=\int_{0}^{x} \frac{1}{1+t^{2}} d t+\int_{-\infty}^{0} \frac{1}{1+t^{2}} d t+\int_{0}^{+\infty} \frac{1}{1+t^{2}} d t \\ &=\int_{-\infty}^{x} \frac{1}{1+t^{2}} d t+\int_{0}^{+\infty} \frac{1}{1+t^{2}} d t \\ &=\int_{\frac{1}{x_{1}}}^{x_{2}} \frac{1}{1+u^{2}} d u+\int_{-x_{1}}^{\frac{1}{x_{1}}} \frac{1}{1+u^{2}} d u \\ &=\int_{-x_{1}}^{x_{2}} \frac{1}{1+u^{2}} d u \\ &=\arctan \left(x_{2}\right)-\arctan \left(-x_{1}\right) \\ &=\arctan \left(x_{2}\right)+\arctan \left(x_{1}\right). \end{aligned}

Hence

\begin{aligned} \tan \left(y_{1}+y_{2}\right) &=\tan \left(y_{1}+y_{2}-\pi\right) \\ &=\tan (\arctan (x)) \\ &=\frac{x_{1}+x_{2}}{1-x_{1} x_{2}} \\ &=\frac{\tan \left(y_{1}\right)+\tan \left(y_{2}\right)}{1-\tan \left(y_{1}\right) \tan \left(y_{2}\right)}. \end{aligned}

The case when $$x_{1}<0$$ may be handled similarly; it then follows that the addition formula holds for all $$y_{1}, y_{2} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) .$$ The case for arbitrary $$y_{1}, y_{2} \in D$$ with $$y_{1}+y_{2} \in D$$ then follows from the periodicity of the tangent function. $$\quad$$ Q.E.D.

This page titled 8.2: The Tangent Function is shared under a CC BY-NC-SA 1.0 license and was authored, remixed, and/or curated by Dan Sloughter via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.