# Reciprocal Trigonometric Functions

> Meet sec, cosec and cot. Their values, their graphs and asymptotes, and the identities that link them.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/reciprocal-trig-functions/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The surveyor's other ratios
A surveyor measures a ramp. She knows $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$, but on this job she needs the ratio the other way up: hypotenuse over opposite, to work out how long a cable must be for each metre of height.

Flipping a trig ratio upside down happens so often that each flipped ratio has its own name: **cosecant**, **secant** and **cotangent**.

**In short:** The reciprocal trig functions are $\operatorname{cosec}\theta = \dfrac{1}{\sin\theta}$, $\sec\theta = \dfrac{1}{\cos\theta}$ and $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$. Their graphs have vertical asymptotes where the original function is zero, and they satisfy $1 + \tan^2 x = \sec^2 x$.

## Starting point: what you already know

- In a right-angled triangle: $\sin\theta = \dfrac{O}{H}$, $\cos\theta = \dfrac{A}{H}$, $\tan\theta = \dfrac{O}{A}$.
- Exact values, such as $\sin 30^\circ = \dfrac{1}{2}$, $\cos 45^\circ = \dfrac{1}{\sqrt{2}}$, $\tan 60^\circ = \sqrt{3}$.
- Radians: $\pi$ radians $= 180^\circ$.
- The graphs of $y = \sin x$, $y = \cos x$ and $y = \tan x$, and the identity $\sin^2 x + \cos^2 x = 1$.
- From the chapter on graphing related functions: zeros of $f$ become vertical asymptotes of $\dfrac{1}{f}$.

## The three new ratios

$$\operatorname{cosec}\theta = \frac{1}{\sin\theta} \qquad \sec\theta = \frac{1}{\cos\theta} \qquad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}$$

In a triangle: $\operatorname{cosec}\theta = \dfrac{H}{O}$, $\sec\theta = \dfrac{H}{A}$, $\cot\theta = \dfrac{A}{O}$.

### Exam tip: Remember the pairs
Look at the third letter: **co**sec goes with **s**in, **sec** goes with **c**os. Each pair has exactly one "co".

### Worked example: Exact values
- $\sec 60^\circ = \dfrac{1}{\cos 60^\circ} = \dfrac{1}{1/2} = 2$
- $\operatorname{cosec} 45^\circ = \dfrac{1}{1/\sqrt{2}} = \sqrt{2}$
- $\cot 30^\circ = \dfrac{1}{\tan 30^\circ} = \sqrt{3}$

### Calculator tip: There is no sec key
Calculators have no sec, cosec or cot keys. Use the reciprocal: $\sec 40^\circ$ is `1 ÷ cos(40)`, or `cos(40)` then the `x⁻¹` key.

To solve $\sec\theta = 2.5$, change it to $\cos\theta = \frac{1}{2.5} = 0.4$, then use `SHIFT cos`.

## The graphs

Use the reciprocal rules: zeros become asymptotes, and heights of $\pm 1$ stay where they are.

%%FIG:sec%%

| | $y = \operatorname{cosec} x$ | $y = \sec x$ | $y = \cot x$ |
|---|---|---|---|
| Asymptotes | where $\sin x = 0$: $x = 0, \pi, 2\pi, \dots$ | where $\cos x = 0$: $x = \tfrac{\pi}{2}, \tfrac{3\pi}{2}, \dots$ | where $\tan x = 0$: $x = 0, \pi, \dots$ |
| Range | $y \le -1$ or $y \ge 1$ | $y \le -1$ or $y \ge 1$ | all real numbers |
| Period | $2\pi$ | $2\pi$ | $\pi$ |
| Symmetry | odd | even | odd |

- $\sec x$ and $\operatorname{cosec} x$ have U-shaped branches that never get between $-1$ and 1.
- $\cot x$ looks like $\tan x$ reflected and shifted, decreasing between each pair of asymptotes. It is zero where $\cos x = 0$.

## New Pythagorean identities

Start from $\sin^2 x + \cos^2 x = 1$.

- Divide every term by $\cos^2 x$: $\tan^2 x + 1 = \sec^2 x$
- Divide every term by $\sin^2 x$: $1 + \cot^2 x = \operatorname{cosec}^2 x$

### Key idea: Three identities
$$\sin^2 x + \cos^2 x = 1 \qquad 1 + \tan^2 x = \sec^2 x \qquad 1 + \cot^2 x = \operatorname{cosec}^2 x$$

**Mind map**

```text
Reciprocal trig
  Definitions
    cosec = 1/sin
    sec = 1/cos
    cot = 1/tan = cos/sin
  Graphs
    Asymptotes at zeros
    sec and cosec: |y| ≥ 1
    cot period π
  Identities
    1 + tan² = sec²
    1 + cot² = cosec²
```

### Year 11: Proving an identity
Show that $\sec x - \cos x = \sin x\tan x$.

Start with the left side:
$$\sec x - \cos x = \frac{1}{\cos x} - \cos x = \frac{1 - \cos^2 x}{\cos x} = \frac{\sin^2 x}{\cos x} = \sin x \cdot \frac{\sin x}{\cos x} = \sin x\tan x$$
Always work on one side only until it matches the other.

### Year 11: Solving an equation
Solve $\sec x = 2$ for $0 \le x \le 2\pi$.

$\cos x = \dfrac{1}{2}$, so $x = \dfrac{\pi}{3}$ or $x = 2\pi - \dfrac{\pi}{3} = \dfrac{5\pi}{3}$.

**Signs in each quadrant.** A reciprocal has the same sign as its partner. If $\tan\theta = \frac{5}{12}$ with $\theta$ in quadrant 3, a 5, 12, 13 triangle gives $\cos\theta = -\frac{12}{13}$, so $\sec\theta = -\frac{13}{12}$.

**Periods after a transformation.** $y = \sec(bx + c)$ and $y = \operatorname{cosec}(bx + c)$ have period $\dfrac{2\pi}{b}$; $y = \cot(bx + c)$ has period $\dfrac{\pi}{b}$.

**Quadratic equations.** Use an identity to get one ratio. $\operatorname{cosec}^2 x = 3 - \cot x$ becomes $1 + \cot^2 x = 3 - \cot x$, so $\cot^2 x + \cot x - 2 = 0$ and $\cot x = 1$ or $-2$. Each value gives two solutions in $[0, 2\pi]$.

**A pair that multiplies to 1.** $(\sec x - \tan x)(\sec x + \tan x) = \sec^2 x - \tan^2 x = 1$. So if $\sec x + \tan x = 3$, then $\sec x - \tan x = \frac{1}{3}$. Adding gives $\sec x = \frac{5}{3}$ and subtracting gives $\tan x = \frac{4}{3}$. The same works for $\operatorname{cosec} x$ and $\cot x$.

**Range of a reciprocal.** If $2 + \sin x$ runs from 1 to 3, then $\dfrac{1}{2 + \sin x}$ runs from $\frac{1}{3}$ to 1.

### Year 12: Where sec² appears in calculus
In Year 12, $\dfrac{d}{dx}\tan x = \sec^2 x$, so $\displaystyle\int \sec^2 x\, dx = \tan x + C$. The identity $\tan^2 x = \sec^2 x - 1$ then lets you integrate $\tan^2 x$.

## Quick summary

- $\operatorname{cosec}\theta = \dfrac{1}{\sin\theta}$, $\sec\theta = \dfrac{1}{\cos\theta}$, $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$.
- Their graphs have vertical asymptotes where the original function is zero.
- $\sec x$ and $\operatorname{cosec} x$ have range $|y| \ge 1$ and period $2\pi$. $\cot x$ has period $\pi$.
- $1 + \tan^2 x = \sec^2 x$ and $1 + \cot^2 x = \operatorname{cosec}^2 x$.

## Quiz

Try these questions from the chapter quiz.

1. $\sec\theta$ is equal to:
   - A. $\dfrac{1}{\sin\theta}$
   - B. $\dfrac{1}{\cos\theta}$
   - C. $\dfrac{1}{\tan\theta}$
   - D. $\dfrac{\sin\theta}{\cos\theta}$
   - Answer: B. $\dfrac{1}{\cos\theta}$. Secant goes with cosine.

2. $\operatorname{cosec}\theta$ in a right triangle is:
   - A. $\dfrac{\text{hypotenuse}}{\text{opposite}}$
   - B. $\dfrac{\text{opposite}}{\text{hypotenuse}}$
   - C. $\dfrac{\text{hypotenuse}}{\text{adjacent}}$
   - D. $\dfrac{\text{adjacent}}{\text{opposite}}$
   - Answer: A. $\dfrac{\text{hypotenuse}}{\text{opposite}}$. It is $\sin\theta$ upside down.

3. Find the exact value of $\sec 60^\circ$.
   - A. $\dfrac{1}{2}$
   - B. 2
   - C. $\dfrac{2}{\sqrt{3}}$
   - D. $\sqrt{3}$
   - Answer: B. 2. $\cos 60^\circ = \dfrac{1}{2}$, flipped is 2.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/reciprocal-trig-functions/#quiz