Reciprocal Trigonometric Functions

Builds on: aqa-alevel-maths-pure-1 / trigonometry — this lesson is the A2-level deep dive.

This lesson covers the three reciprocal trigonometric functions — sec, cosec, and cot — their definitions, graphs, domains, ranges, and the key identities associated with them. These functions extend your trigonometric toolkit and are essential for proving identities, solving equations, and performing integration at A-Level.

Spec Mapping — AQA 7357 Section E Trigonometry. This lesson covers the reciprocal functions sec, cosec, cot with their Pythagorean identities content of Section E, at the depth required for A2-level synoptic questions. Refer to the official AQA specification document for exact wording.

Definitions

The reciprocal trigonometric functions are defined as follows:

\begin{aligned} \sec\theta &= \frac{1}{\cos\theta} \quad \text{(secant)} \\ \csc\theta &= \frac{1}{\sin\theta} \quad \text{(cosecant)} \\ \cot\theta &= \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta} \quad \text{(cotangent)} \end{aligned}

Note: these are reciprocals, not inverses. $\sec\theta \neq \cos^{-1}\theta$ . The inverse function $\arccos$ is a completely different concept.

Domains

Each reciprocal function is undefined where its corresponding function is zero:

Function	Undefined when	Domain exclusions
$\sec\theta$	$\cos\theta = 0$	$\theta \neq \tfrac{\pi}{2} + n\pi$ (i.e., $\theta \neq 90° + 180°n$ )
$\cosec\theta$	$\sin\theta = 0$	$\theta \neq n\pi$ (i.e., $\theta \neq 180°n$ )
$\cot\theta$	$\sin\theta = 0$ (or $\tan\theta = 0$ )	$\theta \neq n\pi$ (i.e., $\theta \neq 180°n$ )

where $n$ is any integer.

Ranges

Function	Range
$\sec\theta$	$\sec\theta \leq -1$ or $\sec\theta \geq 1$ , i.e., $(-\infty, -1] \cup [1, \infty)$
$\cosec\theta$	$\cosec\theta \leq -1$ or $\cosec\theta \geq 1$ , i.e., $(-\infty, -1] \cup [1, \infty)$
$\cot\theta$	All real numbers, $(-\infty, \infty)$

The ranges of $\sec$ and $\cosec$ follow from the fact that $-1 \leq \cos\theta \leq 1$ and $-1 \leq \sin\theta \leq 1$ — taking reciprocals of values between $-1$ and $1$ produces values outside $[-1, 1]$ .

Graphs

The Graph of $y = \sec\theta$

The graph of $\sec\theta$ is obtained by taking the reciprocal of every $y$ -value on the graph of $\cos\theta$ :

Where $\cos\theta = 1$ , $\sec\theta = 1$ (minimum points of $\cos$ become minimum points of $\sec$ ).
Where $\cos\theta = -1$ , $\sec\theta = -1$ (maximum points of $\cos$ become maximum points of $\sec$ ).
Where $\cos\theta = 0$ , $\sec\theta$ is undefined — the graph has vertical asymptotes at $\theta = \tfrac{\pi}{2} + n\pi$ .
The graph consists of U-shaped branches, each sitting between consecutive asymptotes.
Period: $2\pi$ .

The Graph of $y = \cosec\theta$

Similarly:

Vertical asymptotes at $\theta = n\pi$ (where $\sin\theta = 0$ ).
U-shaped branches between consecutive asymptotes.
Where $\sin\theta = 1$ , $\cosec\theta = 1$ (local minima).
Where $\sin\theta = -1$ , $\cosec\theta = -1$ (local maxima).
Period: $2\pi$ .

The Graph of $y = \cot\theta$

Vertical asymptotes at $\theta = n\pi$ (where $\sin\theta = 0$ , i.e., where $\tan\theta = 0$ ).
The graph is a decreasing curve between each pair of asymptotes.
$\cot\theta$ passes through zero where $\tan\theta$ is undefined, i.e., at $\theta = \tfrac{\pi}{2} + n\pi$ .
Period: $\pi$ (not $2\pi$ ).

Key Exact Values

From the standard trigonometric values:

$\theta$	$\sec\theta$	$\cosec\theta$	$\cot\theta$
$0$	$1$	undefined	undefined
$\tfrac{\pi}{6}$	$\tfrac{2}{\sqrt{3}} = \tfrac{2\sqrt{3}}{3}$	$2$	$\sqrt{3}$
$\tfrac{\pi}{4}$	$\sqrt{2}$	$\sqrt{2}$	$1$
$\tfrac{\pi}{3}$	$2$	$\tfrac{2}{\sqrt{3}} = \tfrac{2\sqrt{3}}{3}$	$\tfrac{1}{\sqrt{3}} = \tfrac{\sqrt{3}}{3}$
$\tfrac{\pi}{2}$	undefined	$1$	$0$

Example 1: Find the exact value of $\sec(\tfrac{\pi}{3})$ .

\sec( \pi /3) = 1/\cos( \pi /3) = 1/(1/2) = 2

Example 2: Find the exact value of $\cosec(\tfrac{5\pi}{6})$ .

$\tfrac{5\pi}{6}$ is in the second quadrant, where $\sin$ is positive.

\begin{aligned} \sin(5\pi/6) &= \sin(\pi/6) = \tfrac{1}{2} \\ \csc(5\pi/6) &= \frac{1}{\sin(5\pi/6)} = \frac{1}{1/2} = 2 \end{aligned}

Example 3: Find the exact value of $\cot(\tfrac{2\pi}{3})$ .

$\tfrac{2\pi}{3}$ is in the second quadrant, where $\tan$ (and hence $\cot$ ) is negative.

\begin{aligned} \tan(2 \pi /3) = -\tan( \pi /3) = - \sqrt{3} \\ \cot(2 \pi /3) = 1/(- \sqrt{3}) = -1/ \sqrt{3} = - \sqrt{3}/3 \end{aligned}

The Pythagorean Identities

Identity 1: $\sin^2\theta + \cos^2\theta \equiv 1$

This is the fundamental Pythagorean identity, which you already know.

Identity 2: $1 + \tan^2\theta \equiv \sec^2\theta$

Derivation: Start with $\sin^2\theta + \cos^2\theta \equiv 1$ and divide every term by $\cos^2\theta$ :

\begin{aligned} \frac{\sin^{2}\theta}{\cos^{2}\theta} + \frac{\cos^{2}\theta}{\cos^{2}\theta} &\equiv \frac{1}{\cos^{2}\theta} \\ \tan^{2}\theta + 1 &\equiv \sec^{2}\theta \end{aligned}

Therefore: $1 + \tan^2\theta \equiv \sec^2\theta$

Identity 3: $1 + \cot^2\theta \equiv \cosec^2\theta$

Derivation: Start with $\sin^2\theta + \cos^2\theta \equiv 1$ and divide every term by $\sin^2\theta$ :

\begin{aligned} \frac{\sin^{2}\theta}{\sin^{2}\theta} + \frac{\cos^{2}\theta}{\sin^{2}\theta} &\equiv \frac{1}{\sin^{2}\theta} \\ 1 + \cot^{2}\theta &\equiv \operatorname{cosec}^{2}\theta \end{aligned}

Therefore: $1 + \cot^2\theta \equiv \cosec^2\theta$

These three identities are all forms of the same fundamental relationship. You should be able to derive the second and third from the first, and use all three fluently.

Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Solving Equations with Reciprocal Functions

Example 4: Solve $\sec^2\theta = 4$ for $0 \leq \theta \leq 2\pi$ .

\begin{aligned} \sec^{2} \theta = 4 \\ \cos^{2} \theta = 1/4 \\ \cos \theta = \pm 1/2 \end{aligned}

If $\cos\theta = \tfrac{1}{2}$ : $\theta = \tfrac{\pi}{3}$ or $\theta = \tfrac{5\pi}{3}$ If $\cos\theta = -\tfrac{1}{2}$ : $\theta = \tfrac{2\pi}{3}$ or $\theta = \tfrac{4\pi}{3}$

Solutions: $\theta = \tfrac{\pi}{3}, \tfrac{2\pi}{3}, \tfrac{4\pi}{3}, \tfrac{5\pi}{3}$

Example 5: Solve $2\cosec^2 x - 5\cosec x + 2 = 0$ for $0 < x < 2\pi$ .

Let $u = \cosec x$ :

\begin{aligned} 2u^{2} - 5u + 2 &= 0 \\ (2u - 1)(u - 2) &= 0 \\ u = \tfrac{1}{2} \ &\text{or} \ u = 2 \end{aligned}

If $\cosec x = \tfrac{1}{2}$ , then $\sin x = 2$ . Since $-1 \leq \sin x \leq 1$ , this has no solutions.

If $\cosec x = 2$ , then $\sin x = \tfrac{1}{2}$ . $x = \tfrac{\pi}{6}$ or $x = \tfrac{5\pi}{6}$

Solutions: $x = \tfrac{\pi}{6}, \tfrac{5\pi}{6}$

Example 6: Solve $\cot^2\theta - 3 = 0$ for $0 \leq \theta < 2\pi$ .

\begin{aligned} \cot^{2} \theta = 3 \\ \cot \theta = \pm \sqrt{3} \\ \tan \theta = \pm 1/ \sqrt{3} \end{aligned}

$\tan\theta = \tfrac{1}{\sqrt{3}}$ : $\theta = \tfrac{\pi}{6}, \pi + \tfrac{\pi}{6} = \tfrac{7\pi}{6}$ $\tan\theta = -\tfrac{1}{\sqrt{3}}$ : $\theta = \pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}, 2\pi - \tfrac{\pi}{6} = \tfrac{11\pi}{6}$

Solutions: $\theta = \tfrac{\pi}{6}, \tfrac{5\pi}{6}, \tfrac{7\pi}{6}, \tfrac{11\pi}{6}$

Synoptic Links

This topic connects to:

Proving trig identities (aqa-alevel-maths-trigonometry-depth / proving-trigonometric-identities) — $\sec^2\theta \equiv 1 + \tan^2\theta$ and $\cosec^2\theta \equiv 1 + \cot^2\theta$ are core identity tools.
Integration of $\sec^2 x$ (aqa-alevel-maths-calculus-applications / integration-trigonometric) — $\int \sec^2 x\, dx = \tan x$ is a standard result.
Differentiation (aqa-alevel-maths-calculus-applications / differentiation-trigonometric) — $\tfrac{d}{dx}(\sec x) = \sec x \tan x$ and $\tfrac{d}{dx}(\cot x) = -\cosec^2 x$ .

Summary

$\sec\theta = \tfrac{1}{\cos\theta}$ , $\cosec\theta = \tfrac{1}{\sin\theta}$ , $\cot\theta = \tfrac{\cos\theta}{\sin\theta} = \tfrac{1}{\tan\theta}$ .
Reciprocal functions are undefined where the original function is zero.
$\sec$ and $\cosec$ have range $(-\infty, -1] \cup [1, \infty)$ ; $\cot$ has range $(-\infty, \infty)$ .
Key identities: $1 + \tan^2\theta \equiv \sec^2\theta$ and $1 + \cot^2\theta \equiv \cosec^2\theta$ .
Both identities are derived by dividing $\sin^2\theta + \cos^2\theta \equiv 1$ by $\cos^2\theta$ or $\sin^2\theta$ respectively.
To solve equations involving reciprocal functions, convert to $\sin$ , $\cos$ , or $\tan$ and solve.

Exam Tip: The identities $1 + \tan^2\theta \equiv \sec^2\theta$ and $1 + \cot^2\theta \equiv \cosec^2\theta$ are in the AQA formula booklet. However, you should know how to derive them from $\sin^2\theta + \cos^2\theta \equiv 1$ . When solving equations involving $\cosec$ , $\sec$ , or $\cot$ , it is usually best to convert everything to $\sin$ , $\cos$ , and $\tan$ first.

A-Level Deep Dive: Reciprocal Trigonometric Functions

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section E (Trigonometry), sub-strand E5–E6 (Year 2 content) covers the definitions of secant, cosecant and cotangent and of arcsin, arccos and arctan; their relationships to sine, cosine and tangent; understanding of their graphs; their ranges and domains. Understand and use $\sec^2 \theta = 1 + \tan^2 \theta$ and $\csc^2 \theta = 1 + \cot^2 \theta$ (refer to the official specification document for exact wording). Reciprocal trig is examined in 7357/2 (Pure 2) and feeds 7357/3 (Mechanics & Statistics) anywhere a force decomposition produces $1/\cos\theta$ or $1/\sin\theta$ terms. The AQA formula booklet does list the two derived Pythagorean identities; it does not list the definitions $\sec\theta = 1/\cos\theta$ , $\csc\theta = 1/\sin\theta$ , $\cot\theta = \cos\theta/\sin\theta$ — these are assumed.

Worked example with full mark scheme

Question (8 marks): Solve $\sec^2\theta - 3\tan\theta - 5 = 0$ for $0 \leq \theta < 2\pi$ , giving answers in radians to 3 significant figures.

Solution with mark scheme:

Step 1 — convert to a single trig function.

Using the derived Pythagorean identity $\sec^2\theta = 1 + \tan^2\theta$ :

$1 + \tan^2\theta - 3\tan\theta - 5 = 0$ $\tan^2\theta - 3\tan\theta - 4 = 0$

M1 — correct use of $\sec^2\theta \equiv 1 + \tan^2\theta$ to eliminate $\sec\theta$ . The most common error here is misremembering the identity as $\sec^2\theta = \tan^2\theta - 1$ (sign-flipped) or $\sec^2\theta = 1 - \tan^2\theta$ , both of which destroy the equation.

A1 — correct simplified quadratic in $\tan\theta$ .

Step 2 — factorise the quadratic.

Let $u = \tan\theta$ . Then $u^2 - 3u - 4 = 0$ , which factorises as $(u - 4)(u + 1) = 0$ .

M1 — recognising the hidden quadratic and substituting (or factorising directly in $\tan\theta$ ).

So $\tan\theta = 4$ or $\tan\theta = -1$ .

A1 — both roots correctly identified.

Step 3 — solve $\tan\theta = 4$ on $[0, 2\pi)$ .

Principal value: $\theta = \arctan(4) \approx 1.3258$ rad. Since $\tan\theta$ has period $\pi$ , the second solution in $[0, 2\pi)$ is $\theta \approx 1.3258 + \pi \approx 4.4674$ rad.

M1 — using the periodicity of $\tan\theta$ (period $\pi$ ) to generate the second root.

A1 — $\theta \approx 1.33$ and $\theta \approx 4.47$ to 3 s.f.

Step 4 — solve $\tan\theta = -1$ on $[0, 2\pi)$ .

Principal value: $\theta = -\pi/4$ , outside the required interval. Adding $\pi$ : $\theta = 3\pi/4 \approx 2.356$ rad. Adding another $\pi$ : $\theta = 7\pi/4 \approx 5.498$ rad.

M1 — locating both solutions in $[0, 2\pi)$ from the negative principal value.

A1 — $\theta = 3\pi/4 \approx 2.36$ and $\theta = 7\pi/4 \approx 5.50$ to 3 s.f.

Final answers: $\theta \approx 1.33,\ 2.36,\ 4.47,\ 5.50$ rad.

Total: 8 marks (M4 A4).

Specimen question modelled on the AQA 7357/2 format

Question (6 marks): (a) Prove the identity $\dfrac{1}{\csc\theta - \cot\theta} - \dfrac{1}{\csc\theta + \cot\theta} \equiv 2\cot\theta$ . (4) (b) Hence find the exact value of the expression when $\theta = \pi/3$ . (2)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — combining the two fractions over a common denominator $(\csc\theta - \cot\theta)(\csc\theta + \cot\theta)$ .
A1 (AO1.1b) — numerator simplifies to $(\csc\theta + \cot\theta) - (\csc\theta - \cot\theta) = 2\cot\theta$ .
M1 (AO2.1) — denominator: $(\csc\theta)^2 - (\cot\theta)^2 = \csc^2\theta - \cot^2\theta$ , which by the derived Pythagorean identity $1 + \cot^2\theta = \csc^2\theta$ equals $1$ .
A1 (AO2.5) — concluding $\dfrac{2\cot\theta}{1} = 2\cot\theta$ as required.

(b)

M1 (AO1.1b) — $\cot(\pi/3) = \cos(\pi/3)/\sin(\pi/3) = (1/2)/(\sqrt{3}/2) = 1/\sqrt{3}$ .
A1 (AO1.1b) — $2\cot(\pi/3) = 2/\sqrt{3} = 2\sqrt{3}/3$ (rationalised).

Total: 6 marks split AO1 = 4, AO2 = 2. AQA 7357 reciprocal-trig identity proofs reward the explicit invocation of $\csc^2\theta - \cot^2\theta = 1$ — without naming this step you cannot earn the AO2 mark.

Synoptic links

Connects to:

Section E1 — basic trigonometry: every reciprocal definition rests on the parent function. $\sec\theta$ inherits the period of $\cos\theta$ (period $2\pi$ ) and is undefined wherever $\cos\theta = 0$ (at $\theta = \pi/2 + k\pi$ ). Mastery of where $\cos\theta$ , $\sin\theta$ and $\tan\theta$ vanish is a prerequisite for sketching the reciprocal graphs.
Section G — exponentials and logarithms: identities like $\sec\theta + \tan\theta = e^x$ appear in Year 2 hyperbolic contexts and in integration by Weierstrass substitution ( $t = \tan(\theta/2)$ ). Algebraic confidence with reciprocal-trig manipulations transfers directly.
Section J — differentiation: $\frac{d}{dx}(\tan x) = \sec^2 x$ , $\frac{d}{dx}(\sec x) = \sec x \tan x$ , $\frac{d}{dx}(\cot x) = -\csc^2 x$ , $\frac{d}{dx}(\csc x) = -\csc x \cot x$ . These four derivatives are listed in the formula booklet; integrating them in reverse requires you to recognise the reciprocal-trig structure.
Section K — integration: $\int \sec^2 x \, dx = \tan x + C$ and $\int \sec x \tan x \, dx = \sec x + C$ are directly reciprocal-trig. Harder integrals like $\int \sec^3 x \, dx$ use integration by parts plus the identity $\sec^2 x = 1 + \tan^2 x$ to reduce powers.
Section O — Mechanics (resolving forces on inclined planes): the reaction force on a slope of angle $\alpha$ often appears as $N = mg/\cos\alpha = mg\sec\alpha$ . Friction analyses with limiting equilibrium produce $\tan\alpha = \mu$ — i.e. the coefficient of friction is the cotangent's reciprocal. Reciprocal-trig fluency speeds these problems.

Reciprocal Trigonometric Functions

Reciprocal Trigonometric Functions

Definitions

Domains

Ranges

Graphs

The Graph of $y = \sec\theta$

The Graph of $y = \cosec\theta$

The Graph of $y = \cot\theta$

Key Exact Values

The Pythagorean Identities

Identity 1: $\sin^2\theta + \cos^2\theta \equiv 1$

Identity 2: $1 + \tan^2\theta \equiv \sec^2\theta$

Identity 3: $1 + \cot^2\theta \equiv \cosec^2\theta$

Specimen Question (Modelled on the AQA Paper Format)

Solving Equations with Reciprocal Functions

Synoptic Links

Summary

A-Level Deep Dive: Reciprocal Trigonometric Functions

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357/2 format

Synoptic links

Mark-scheme literacy

More in Mathematics

Reciprocal Trigonometric Functions

Reciprocal Trigonometric Functions

Definitions

Domains

Ranges

Graphs

The Graph of y=sec⁡θy = \sec\thetay=secθ

The Graph of y=cosec⁡θy = \cosec\thetay=cosecθ

The Graph of y=cot⁡θy = \cot\thetay=cotθ

Key Exact Values

The Pythagorean Identities

Identity 1: sin⁡2θ+cos⁡2θ≡1\sin^2\theta + \cos^2\theta \equiv 1sin2θ+cos2θ≡1

Identity 2: 1+tan⁡2θ≡sec⁡2θ1 + \tan^2\theta \equiv \sec^2\theta1+tan2θ≡sec2θ

Identity 3: 1+cot⁡2θ≡cosec⁡2θ1 + \cot^2\theta \equiv \cosec^2\theta1+cot2θ≡cosec2θ

Specimen Question (Modelled on the AQA Paper Format)

Solving Equations with Reciprocal Functions

Synoptic Links

Summary

A-Level Deep Dive: Reciprocal Trigonometric Functions

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357/2 format

Synoptic links

Mark-scheme literacy

More in Mathematics

The Graph of $y = \sec\theta$

The Graph of $y = \cosec\theta$

The Graph of $y = \cot\theta$

Identity 1: $\sin^2\theta + \cos^2\theta \equiv 1$

Identity 2: $1 + \tan^2\theta \equiv \sec^2\theta$

Identity 3: $1 + \cot^2\theta \equiv \cosec^2\theta$