Reciprocal Trigonometric Functions

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.

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Definitions

The reciprocal trigonometric functions are defined as follows:

sec θ = 1/cos θ       (secant)
cosec θ = 1/sin θ     (cosecant)
cot θ = 1/tan θ = cos θ/sin θ   (cotangent)

Note: these are reciprocals, not inverses. sec θ ≠ cos⁻¹ θ. The inverse function arccos is a completely different concept.

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Domains

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

Function	Undefined when	Domain exclusions
sec θ	cos θ = 0	θ ≠ π/2 + nπ (i.e., θ ≠ 90° + 180°n)
cosec θ	sin θ = 0	θ ≠ nπ (i.e., θ ≠ 180°n)
cot θ	sin θ = 0 (or tan θ = 0)	θ ≠ nπ (i.e., θ ≠ 180°n)

where n is any integer.

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Ranges

Function	Range
sec θ	sec θ ≤ −1 or sec θ ≥ 1, i.e., (−∞, −1] ∪ [1, ∞)
cosec θ	cosec θ ≤ −1 or cosec θ ≥ 1, i.e., (−∞, −1] ∪ [1, ∞)
cot θ	All real numbers, (−∞, ∞)

The ranges of sec and cosec follow from the fact that −1 ≤ cos θ ≤ 1 and −1 ≤ sin θ ≤ 1 — taking reciprocals of values between −1 and 1 produces values outside [−1, 1].

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Graphs

The Graph of y = sec θ

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

• Where cos θ = 1, sec θ = 1 (minimum points of cos become minimum points of sec).
• Where cos θ = −1, sec θ = −1 (maximum points of cos become maximum points of sec).
• Where cos θ = 0, sec θ is undefined — the graph has vertical asymptotes at θ = π/2 + nπ.
• The graph consists of U-shaped branches, each sitting between consecutive asymptotes.
• Period: 2π.

The Graph of y = cosec θ

Similarly:

• Vertical asymptotes at θ = nπ (where sin θ = 0).
• U-shaped branches between consecutive asymptotes.
• Where sin θ = 1, cosec θ = 1 (local minima).
• Where sin θ = −1, cosec θ = −1 (local maxima).
• Period: 2π.

The Graph of y = cot θ

• Vertical asymptotes at θ = nπ (where sin θ = 0, i.e., where tan θ = 0).
• The graph is a decreasing curve between each pair of asymptotes.
• cot θ passes through zero where tan θ is undefined, i.e., at θ = π/2 + nπ.
• Period: π (not 2π).

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Key Exact Values

From the standard trigonometric values:

θ	sec θ	cosec θ	cot θ
0	1	undefined	undefined
π/6	2/√3 = 2√3/3	2	√3
π/4	√2	√2	1
π/3	2	2/√3 = 2√3/3	1/√3 = √3/3
π/2	undefined	1	0

Example 1: Find the exact value of sec(π/3).

sec(π/3) = 1/cos(π/3) = 1/(1/2) = 2

Example 2: Find the exact value of cosec(5π/6).

5π/6 is in the second quadrant, where sin is positive.

sin(5π/6) = sin(π/6) = 1/2
cosec(5π/6) = 1/sin(5π/6) = 1/(1/2) = 2

Example 3: Find the exact value of cot(2π/3).

2π/3 is in the second quadrant, where tan (and hence cot) is negative.

tan(2π/3) = −tan(π/3) = −√3
cot(2π/3) = 1/(−√3) = −1/√3 = −√3/3

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The Pythagorean Identities

Identity 1: sin²θ + cos²θ ≡ 1

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

Identity 2: 1 + tan²θ ≡ sec²θ

Derivation: Start with sin²θ + cos²θ ≡ 1 and divide every term by cos²θ:

sin²θ/cos²θ + cos²θ/cos²θ ≡ 1/cos²θ
tan²θ + 1 ≡ sec²θ

Therefore: 1 + tan²θ ≡ sec²θ

Identity 3: 1 + cot²θ ≡ cosec²θ

Derivation: Start with sin²θ + cos²θ ≡ 1 and divide every term by sin²θ:

sin²θ/sin²θ + cos²θ/sin²θ ≡ 1/sin²θ
1 + cot²θ ≡ cosec²θ

Therefore: 1 + cot²θ ≡ cosec²θ

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.

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Solving Equations with Reciprocal Functions

Example 4: Solve sec²θ = 4 for 0 ≤ θ ≤ 2π.

sec²θ = 4
cos²θ = 1/4
cos θ = ±1/2

If cos θ = 1/2: θ = π/3 or θ = 5π/3 If cos θ = −1/2: θ = 2π/3 or θ = 4π/3

Solutions: θ = π/3, 2π/3, 4π/3, 5π/3

Example 5: Solve 2cosec²x − 5cosec x + 2 = 0 for 0 < x < 2π.

Let u = cosec x:

2u² − 5u + 2 = 0
(2u − 1)(u − 2) = 0
u = 1/2 or u = 2

If cosec x = 1/2, then sin x = 2. Since −1 ≤ sin x ≤ 1, this has no solutions.

If cosec x = 2, then sin x = 1/2. x = π/6 or x = 5π/6

Solutions: x = π/6, 5π/6

Example 6: Solve cot²θ − 3 = 0 for 0 ≤ θ < 2π.

cot²θ = 3
cot θ = ±√3
tan θ = ±1/√3

tan θ = 1/√3: θ = π/6, π + π/6 = 7π/6 tan θ = −1/√3: θ = π − π/6 = 5π/6, 2π − π/6 = 11π/6

Solutions: θ = π/6, 5π/6, 7π/6, 11π/6

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Summary

• sec θ = 1/cos θ, cosec θ = 1/sin θ, cot θ = cos θ/sin θ = 1/tan θ.
• Reciprocal functions are undefined where the original function is zero.
• sec and cosec have range (−∞, −1] ∪ [1, ∞); cot has range (−∞, ∞).
• Key identities: 1 + tan²θ ≡ sec²θ and 1 + cot²θ ≡ cosec²θ.
• Both identities are derived by dividing sin²θ + cos²θ ≡ 1 by cos²θ or sin²θ respectively.
• To solve equations involving reciprocal functions, convert to sin, cos, or tan and solve.

Exam Tip: The identities 1 + tan²θ ≡ sec²θ and 1 + cot²θ ≡ cosec²θ are in the AQA formula booklet. However, you should know how to derive them from sin²θ + cos²θ ≡ 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 $\sec^2 \theta = 1 + \tan^2 \theta$sec2θ=1+tan2θ and $\csc^2 \theta = 1 + \cot^2 \theta$csc2θ=1+cot2θ (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$1/cosθ or $1/\sin\theta$1/sinθ terms. The AQA formula booklet does list the two derived Pythagorean identities; it does not list the definitions $\sec\theta = 1/\cos\theta$secθ=1/cosθ, $\csc\theta = 1/\sin\theta$cscθ=1/sinθ, $\cot\theta = \cos\theta/\sin\theta$cotθ=cosθ/sinθ — these are assumed.

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Worked example with full mark scheme

Question (8 marks): Solve $\sec^2\theta - 3\tan\theta - 5 = 0$sec2θ−3tanθ−5=0 for $0 \leq \theta < 2\pi$0≤θ<2π, 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 $\sec^2\theta = 1 + \tan^2\theta$sec2θ=1+tan2θ:

$1 + \tan^2\theta - 3\tan\theta - 5 = 0$1+tan2θ−3tanθ−5=0 $\tan^2\theta - 3\tan\theta - 4 = 0$tan2θ−3tanθ−4=0

M1 — correct use of $\sec^2\theta \equiv 1 + \tan^2\theta$sec2θ≡1+tan2θ to eliminate $\sec\theta$secθ. The most common error here is misremembering the identity as $\sec^2\theta = \tan^2\theta - 1$sec2θ=tan2θ−1 (sign-flipped) or $\sec^2\theta = 1 - \tan^2\theta$sec2θ=1−tan2θ, both of which destroy the equation.

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

Step 2 — factorise the quadratic.

Let $u = \tan\theta$u=tanθ. Then $u^2 - 3u - 4 = 0$u2−3u−4=0, which factorises as $(u - 4)(u + 1) = 0$(u−4)(u+1)=0.

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

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

A1 — both roots correctly identified.

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

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

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

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

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

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

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

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

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

Total: 8 marks (M4 A4).

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Specimen question modelled on the AQA 7357/2 format

Question (6 marks): (a) Prove the identity $\dfrac{1}{\csc\theta - \cot\theta} - \dfrac{1}{\csc\theta + \cot\theta} \equiv 2\cot\theta$cscθ−cotθ1​−cscθ+cotθ1​≡2cotθ. (4) (b) Hence find the exact value of the expression when $\theta = \pi/3$θ=π/3. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — combining the two fractions over a common denominator $(\csc\theta - \cot\theta)(\csc\theta + \cot\theta)$(cscθ−cotθ)(cscθ+cotθ).
• A1 (AO1.1b) — numerator simplifies to $(\csc\theta + \cot\theta) - (\csc\theta - \cot\theta) = 2\cot\theta$(cscθ+cotθ)−(cscθ−cotθ)=2cotθ.
• M1 (AO2.1) — denominator: $(\csc\theta)^2 - (\cot\theta)^2 = \csc^2\theta - \cot^2\theta$(cscθ)2−(cotθ)2=csc2θ−cot2θ, which by the derived Pythagorean identity $1 + \cot^2\theta = \csc^2\theta$1+cot2θ=csc2θ equals $1$1.
• A1 (AO2.5) — concluding $\dfrac{2\cot\theta}{1} = 2\cot\theta$12cotθ​=2cotθ as required.

(b)

• M1 (AO1.1b) — $\cot(\pi/3) = \cos(\pi/3)/\sin(\pi/3) = (1/2)/(\sqrt{3}/2) = 1/\sqrt{3}$cot(π/3)=cos(π/3)/sin(π/3)=(1/2)/(3​/2)=1/3​.
• A1 (AO1.1b) — $2\cot(\pi/3) = 2/\sqrt{3} = 2\sqrt{3}/3$2cot(π/3)=2/3​=23​/3 (rationalised).

Total: 6 marks split AO1 = 4, AO2 = 2. AQA 7357 reciprocal-trig identity proofs reward the explicit invocation of $\csc^2\theta - \cot^2\theta = 1$csc2θ−cot2θ=1 — without naming this step you cannot earn the AO2 mark.

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Synoptic links

Connects to:

1. Section E1 — basic trigonometry: every reciprocal definition rests on the parent function. $\sec\theta$secθ inherits the period of $\cos\theta$cosθ (period $2\pi$2π) and is undefined wherever $\cos\theta = 0$cosθ=0 (at $\theta = \pi/2 + k\pi$θ=π/2+kπ). Mastery of where $\cos\theta$cosθ, $\sin\theta$sinθ and $\tan\theta$tanθ vanish is a prerequisite for sketching the reciprocal graphs.

2. Section G — exponentials and logarithms: identities like $\sec\theta + \tan\theta = e^x$secθ+tanθ=ex appear in Year 2 hyperbolic contexts and in integration by Weierstrass substitution ($t = \tan(\theta/2)$t=tan(θ/2)). Algebraic confidence with reciprocal-trig manipulations transfers directly.

3. Section J — differentiation: $\frac{d}{dx}(\tan x) = \sec^2 x$dxd​(tanx)=sec2x, $\frac{d}{dx}(\sec x) = \sec x \tan x$dxd​(secx)=secxtanx, $\frac{d}{dx}(\cot x) = -\csc^2 x$dxd​(cotx)=−csc2x, $\frac{d}{dx}(\csc x) = -\csc x \cot x$dxd​(cscx)=−cscxcotx. These four derivatives are listed in the formula booklet; integrating them in reverse requires you to recognise the reciprocal-trig structure.

4. Section K — integration: $\int \sec^2 x \, dx = \tan x + C$∫sec2xdx=tanx+C and $\int \sec x \tan x \, dx = \sec x + C$∫secxtanxdx=secx+C are directly reciprocal-trig. Harder integrals like $\int \sec^3 x \, dx$∫sec3xdx use integration by parts plus the identity $\sec^2 x = 1 + \tan^2 x$sec2x=1+tan2x to reduce powers.

5. Section O — Mechanics (resolving forces on inclined planes): the reaction force on a slope of angle $\alpha$α often appears as $N = mg/\cos\alpha = mg\sec\alpha$N=mg/cosα=mgsecα. 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.

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Mark-scheme literacy

Reciprocal-trig questions on AQA 7357/2 split AO marks roughly: