Reciprocal Trigonometry Functions

A2 · P3 · 3 min

Secant, cosecant and cotangent are the reciprocals of cosine, sine and tangent. They bring two new Pythagorean identities, and their graphs have vertical asymptotes wherever the original function is zero.

Definitions and graphs

Key result

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

$y = \sec x$ : asymptotes where $\cos x = 0$ ( $x = \pm 90^\circ, \pm 270^\circ, \ldots$ ); range $y \leq -1$ or $y \geq 1$ ; period $360^\circ$ .
$y = \operatorname{cosec} x$ : asymptotes where $\sin x = 0$ ( $x = 0, \pm 180^\circ, \ldots$ ); same range; period $360^\circ$ .
$y = \cot x$ : asymptotes where $\sin x = 0$ ; range all real numbers; period $180^\circ$ ; decreasing on each branch.

y = 1 / cos x y = cos x

The U-shaped pieces of $\sec x$ sit on the peaks and troughs of $\cos x$ : where $\cos x = 1$ , $\sec x = 1$ ; as $\cos x \to 0$ , $\sec x \to \pm\infty$ .

The Pythagorean identities

Key result

1 + \tan^2\theta \equiv \sec^2\theta, \qquad 1 + \cot^2\theta \equiv \operatorname{cosec}^2\theta

Proof

Divide $\sin^2\theta + \cos^2\theta \equiv 1$ by $\cos^2\theta$ : $\tan^2\theta + 1 \equiv \sec^2\theta$ . Divide by $\sin^2\theta$ instead: $1 + \cot^2\theta \equiv \operatorname{cosec}^2\theta$ .

Exact values

Find the exact values of $\sec 150^\circ$ , $\operatorname{cosec}\tfrac{\pi}{4}$ and $\cot 300^\circ$ .

Solution

$\cos 150^\circ = -\tfrac{\sqrt{3}}{2}$ , so $\sec 150^\circ = -\tfrac{2}{\sqrt{3}}$ .

$\sin\tfrac{\pi}{4} = \tfrac{\sqrt{2}}{2}$ , so $\operatorname{cosec}\tfrac{\pi}{4} = \sqrt{2}$ .

$\tan 300^\circ = -\sqrt{3}$ , so $\cot 300^\circ = -\tfrac{1}{\sqrt{3}}$ .

From one function to another

Given that $\sec\theta = 3$ and $\theta$ is acute, find the exact values of $\tan\theta$ and $\sin\theta$ .

Solution

$\tan^2\theta = \sec^2\theta - 1 = 8$ , so $\tan\theta = 2\sqrt{2}$ (positive because $\theta$ is acute).

$\cos\theta = \tfrac{1}{3}$ and $\sin\theta = \tan\theta\cos\theta = \tfrac{2\sqrt{2}}{3}$ .

Solving equations

Convert to a single function, usually $\tan$ or $\cot$ via the Pythagorean identities, or to $\sin$ or $\cos$ via the definitions.

A quadratic in tan

Solve $\sec^2\theta - 3\tan\theta - 5 = 0$ for $0^\circ \leq \theta \leq 360^\circ$ .

Solution

$1 + \tan^2\theta - 3\tan\theta - 5 = 0 \Rightarrow \tan^2\theta - 3\tan\theta - 4 = 0 \Rightarrow (\tan\theta - 4)(\tan\theta + 1) = 0$ .

$\tan\theta = 4$ : $\theta = 76.0^\circ, 256.0^\circ$ . $\tan\theta = -1$ : $\theta = 135^\circ, 315^\circ$ .

Cosec and cot

Solve $\operatorname{cosec}^2\theta + 3\cot\theta = 5$ for $0^\circ < \theta < 180^\circ$ .

Solution

$1 + \cot^2\theta + 3\cot\theta - 5 = 0 \Rightarrow \cot^2\theta + 3\cot\theta - 4 = 0 \Rightarrow (\cot\theta + 4)(\cot\theta - 1) = 0$ .

$\cot\theta = 1 \Rightarrow \tan\theta = 1 \Rightarrow \theta = 45^\circ$ . $\cot\theta = -4 \Rightarrow \tan\theta = -\tfrac{1}{4} \Rightarrow \theta = 166.0^\circ$ .

Using the definitions directly

Solve $2\sin\theta = \operatorname{cosec}\theta$ for $0 \leq \theta \leq 2\pi$ .

Solution

$2\sin\theta = \dfrac{1}{\sin\theta} \Rightarrow \sin^2\theta = \tfrac{1}{2} \Rightarrow \sin\theta = \pm\tfrac{1}{\sqrt{2}}$ .

$\theta = \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4}$ .

Watch out

$\sec^{-1}$ on a calculator does not exist as a button. To solve $\sec\theta = 3$ , rewrite as $\cos\theta = \tfrac{1}{3}$ . Likewise $\cot\theta = 2$ is $\tan\theta = \tfrac{1}{2}$ .

Exam tip

Proofs using these identities are marked on the flow of steps. Write " $\equiv$ " between expressions, keep one side fixed, and state the identity you use the first time it appears.

Practice

Question

Find the exact values of $\operatorname{cosec} 210^\circ$ and $\cot\tfrac{5\pi}{6}$ .
Given $\cot\theta = \tfrac{3}{4}$ and $\theta$ is reflex, find $\sin\theta$ and $\sec\theta$ .
Solve $\tan^2\theta = 2\sec\theta - 1$ for $0^\circ \leq \theta \leq 360^\circ$ .
Prove that $\sec^2\theta + \operatorname{cosec}^2\theta \equiv \sec^2\theta\operatorname{cosec}^2\theta$ .

Answers

$-2$ ; $-\sqrt{3}$ .
Reflex with positive $\cot$ means third quadrant: $\sin\theta = -\tfrac{4}{5}$ , $\cos\theta = -\tfrac{3}{5}$ , $\sec\theta = -\tfrac{5}{3}$ .
$\sec^2\theta - 2\sec\theta = 0 \Rightarrow \sec\theta = 2$ (not $0$ ): $\cos\theta = \tfrac{1}{2}$ , $\theta = 60^\circ, 300^\circ$ .
$\dfrac{1}{\cos^2} + \dfrac{1}{\sin^2} = \dfrac{\sin^2 + \cos^2}{\sin^2\cos^2} = \dfrac{1}{\sin^2\theta\cos^2\theta}$ .

Definitions and graphs

The Pythagorean identities

Solving equations

Practice

Console