8.2: The Tangent Function

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 $\PageIndex{3}$

(Addition formula for tangent)

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.