# Reciprocal Trigonometric Functions: cosec(x), sec(x), cotan(x)

When we take 1 divided by each of the original trigonometric functions we obtain the **reciprocal trigonometric functions**:

$\text{cosec}(x)=\frac{1}{\sin(x)},\hspace{30pt}\text{sec}(x)=\frac{1}{\cos(x)}, \hspace{30pt}\text{cot}(x)=\frac{1}{\tan(x)}$.

An easy way to remember which one is which is to look at the 3rd letter in each reciprocal function. For example, the third letter of co**s**ec is **s** and so this is one over **s**in. We can solve equations involving reciprocal trigonometric functions in the same way we solve trigonometric equations. See Example 1. You might also notice that $\text{cot}(x)=\frac{1}{\tan(x)}=\frac{\cos(x)}{\sin(x)}$.

These reciprocal functions can be evaluated for angles given in degrees and radians. For example, $\text{cosec}(270^\circ)=\frac{1}{\sin(270^\circ)}=\frac{1}{-1}=-1$ or $\text{sec}(2\pi)=\frac{1}{\cos(2\pi)}=\frac{1}{1}=1$. However, like the graph of $\tan(x)$, all of the reciprocal trigonometric functions have periodic asymptotes. We cannot evaluate them when the denominator is 0. See graphs below.

## Graphs of the Reciprocal Trigonometric Functions

Here are the graphs of the $\text{cosec}(x)$, $\text{sec}(x)$ and $\text{cotan}(x)$ respectively. As you can see the angles are given in radians:

Consider the graph of $\text{cosec}(x)$, for example. It is defined by $\frac{1}{\sin(x)}$ and so has asymptotes whenever $\sin(x)=0$. As you can see from the graph, this occurs for multiples of $\pi$ i.e, when $x=n\pi, n\in{\mathbb Z}$. As $x$ tends towards these values, the curve grows infinitely positive if $\sin(x)$ is positive and infinitely negative if $\sin(x)$ is negative. This asymptotic behaviour can be seen in the other two curves as well.

Now consider the graph of $\text{sec}(x)$. This graph has periodic maxima and minima at multiples of $\pi$. For example, there is a maximum at $(\pi,-1)$ since $\text{sec}(\pi)=\frac{1}{\cos(\pi)}=\frac{1}{-1}=-1$. Notice that $\cos(x)$ takes values between $-1$ and $1$ and so $\frac{1}{\cos(x)}$ will take values less than $-1$ or greater than $1$. This follows from the fact that dividing by a number less than 1 gives a result greater than 1, etc. Hence, the domain and range for $\text{sec}(x)$ are:

DOMAIN of $\text{sec}(x):\hspace{30pt} \left\lbrace x\in {\mathbb R}: x\neq \frac{(2n+1)\pi}{2},n\in{\mathbb Z}\right\rbrace$

RANGE of $\text{sec}(x): \hspace{30pt}\left\lbrace y\in {\mathbb R}:y\leq-1,y\geq 1\right\rbrace$

where we note that we cannot evaluate $\cos(x)$ nor $\text{sec}(x)$ at odd multiples of $\frac{\pi}{2}$. See more on domain and range. The domain of both $\text{cosec(x)}$ and $\text{cotan}(x)$ is $\left\lbrace x\in {\mathbb R}: x\neq n\pi,n\in{\mathbb Z}\right\rbrace$. The range of $\text{cosec(x)}$ is the same as that of $\text{sec(x)}$. Since the range of $\tan(x)$ is $y\in{\mathbb R}$ with no exceptions, the range of $\text{cotan}(x)$ is also $y\in{\mathbb R}$. See Example 2 for a transformation example.

## Reciprocal Trigonometric Function Identities

There are two more trigonometric identities you should learn. See other trigonometric identities. These two new identities are useful when solving certain trigonometric equations:

$1+\tan^2(x)=\text{sec}^2(x)$

$1+\text{cot}^2(x)=\text{cosec}^2(x)$

See Example 3. There is a quirky way to remember these identities. ‘The one with the tan is sexy’ and ‘the one with the coat is cosy’. They both come from the trigonometric identity $\sin^2(x)+\cos^2(x)=1$. Dividing both sides by $\cos^2(x)$ gives the first identity and by $\sin^2(x)$ gives the second.

## Questions by Topic

## Examples

Solve $5\text{cosec}(2x)=10$ for $-2\pi<x<2\pi$.

By dividing both sides by 5 and writing $\text{cosec}(2x)$ as $\frac{1}{\sin(2x)}$ we get the equation:

$\frac{1}{\sin(2x)}=2$

which can be rearranged to $\sin(2x)=\frac{1}{2}$. Let $\theta=2x$ so that we are looking for solutions to $\sin(\theta)=\frac{1}{2}$ for $-4\pi<\theta<4\pi$. See more on solving trigonometric equations. The $\theta$ solutions are:

$\theta=\frac{-23}{6}\pi,\frac{-19}{6}\pi,\frac{-11}{6}\pi, \frac{-7}{6}\pi,\frac{1}{6}\pi, \frac{5}{6}\pi, \frac{13}{6}\pi, \frac{17}{6}\pi$

and so the $x$ solutions are

$x=\frac{-23}{12}\pi,\frac{-19}{12}\pi,\frac{-11}{12}\pi, \frac{-7}{12}\pi,\frac{1}{12}\pi, \frac{5}{12}\pi, \frac{13}{12}\pi, \frac{17}{12}\pi$

Sketch and state the domain and range of the graph of $y=\text{cosec}(2x)+1$.

Recall from the transformations page that replacing $x$ with $2x$ shrinks the original graph by half in the $x$-direction. Adding 1 translates the graph upwards by 1. Hence the graph of $y=\text{cosec}(2x)+1$ is as follows:

The domain of $y=\text{cosec}(2x)+1$ is $\left\lbrace x\in {\mathbb R}: x\neq \frac{n\pi}{2},n\in{\mathbb Z}\right\rbrace$. The range is $\left\lbrace y\in {\mathbb R}:y\leq 0,y\geq 2\right\rbrace$.

- Given that $\tan(\theta)=-\frac{3}{4}$ and $\theta$ is obtuse, find the value of $\sec(\theta)$.
- Show that $\text{sec}^2(\theta)+\text{cosec}^2(\theta)\equiv\text{cosec}^2(\theta)\text{sec}^2(\theta)$

- Using the identity $1+\tan^2(\theta)=\sec^2(\theta)$ we have $\sec^2(\theta)=1+\left(-\frac{3}{4}\right)^2=\frac{25}{16}$. Hence, $\sec(\theta)=\pm \frac{5}{4}$. $\cos(\theta)$ is negative when $\theta$ is obtuse and so $\sec(\theta)=-\frac{5}{4}$.
- Using the definitions: $\begin{array}{l}\text{sec}^2(\theta)+\text{cosec}^2(\theta)&\equiv&\frac{1}{\cos^2(\theta)}+\frac{1}{\sin^2(\theta)}\\&\equiv&\frac{\sin^2(\theta)}{\sin^2(\theta)\cos^2(\theta)}+\frac{\cos^2(\theta)}{\sin^2(\theta)\cos^2(\theta)}\\&\equiv&\frac{\sin^2(\theta)+\cos^2(\theta)}{\sin^2(\theta)\cos^2(\theta)}\\&\equiv&\frac{1}{\sin^2(\theta)\cos^2(\theta)}\\&\equiv&\frac{1}{\sin^2(\theta)}\times\frac{1}{\cos^2(\theta)}\\&\equiv&\text{cosec}^2(\theta)\text{sec}^2(\theta)\end{array}$