Reciprocal Trigonometric Functions

The reciprocal trigonometric functions — secant (sec), cosecant (cosec) and cotangent (cot) — are the reciprocals of cosine, sine and tangent respectively. They have their own identities and graphs, and are tested at A-Level (9MA0).

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Definitions

Function	Definition	Undefined when
sec(theta)	1/cos(theta)	cos(theta) = 0, i.e. theta = pi/2 + npi
cosec(theta)	1/sin(theta)	sin(theta) = 0, i.e. theta = npi
cot(theta)	1/tan(theta) = cos(theta)/sin(theta)	sin(theta) = 0, i.e. theta = npi

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Key Identities

Identity 1

1 + tan²(theta) = sec²(theta)

Derived from sin² + cos² = 1 by dividing by cos²:

sin²/cos² + cos²/cos² = 1/cos²

tan² + 1 = sec²

Identity 2

1 + cot²(theta) = cosec²(theta)

Derived from sin² + cos² = 1 by dividing by sin²:

sin²/sin² + cos²/sin² = 1/sin²

1 + cot² = cosec²

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Graphs

y = sec(x)

• Period: 2pi
• Vertical asymptotes: x = pi/2 + npi (where cos(x) = 0)
• Range: y <= -1 or y >= 1 (never between -1 and 1)
• The graph consists of U-shaped and inverted-U-shaped branches
• sec(0) = 1, sec(pi) = -1

y = cosec(x)

• Period: 2pi
• Vertical asymptotes: x = npi (where sin(x) = 0)
• Range: y <= -1 or y >= 1
• cosec(pi/2) = 1, cosec(3pi/2) = -1

y = cot(x)

• Period: pi
• Vertical asymptotes: x = npi (where sin(x) = 0)
• Range: all real numbers
• Decreasing on each branch
• cot(pi/4) = 1, cot(pi/2) = 0, cot(3pi/4) = -1

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Exact Values

Angle	sec	cosec	cot
pi/6	2/sqrt(3) = 2sqrt(3)/3	2	sqrt(3)
pi/4	sqrt(2)	sqrt(2)	1
pi/3	2	2/sqrt(3) = 2sqrt(3)/3	1/sqrt(3) = sqrt(3)/3

Worked Example 1

Find the exact value of sec(pi/3).

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

Worked Example 2

Find the exact value of cosec(5pi/6).

sin(5pi/6) = sin(pi - 5pi/6) = sin(pi/6) = 1/2

cosec(5pi/6) = 1/(1/2) = 2

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Solving Equations

Worked Example 3

Solve sec²(x) = 4 for 0 <= x < 2pi.

1/cos²(x) = 4

cos²(x) = 1/4

cos(x) = 1/2 or cos(x) = -1/2

cos(x) = 1/2: x = pi/3, 5pi/3

cos(x) = -1/2: x = 2pi/3, 4pi/3

Solutions: x = pi/3, 2pi/3, 4pi/3, 5pi/3

Worked Example 4

Solve 2cot²(x) + 5cosec(x) = 1 for 0 < x < 2pi.

Using 1 + cot²(x) = cosec²(x), so cot²(x) = cosec²(x) - 1:

2(cosec²(x) - 1) + 5cosec(x) = 1

2cosec²(x) - 2 + 5cosec(x) = 1

2cosec²(x) + 5cosec(x) - 3 = 0

Let c = cosec(x):

2c² + 5c - 3 = 0

(2c - 1)(c + 3) = 0

c = 1/2 or c = -3

cosec(x) = 1/2: sin(x) = 2, which is impossible (|sin(x)| <= 1)

cosec(x) = -3: sin(x) = -1/3

x = pi + arcsin(1/3) ≈ pi + 0.3398 ≈ 3.481

x = 2pi - arcsin(1/3) ≈ 2pi - 0.3398 ≈ 5.943

Solutions: x ≈ 3.481 and x ≈ 5.943

Worked Example 5

Solve sec(x) = 2cos(x) + tan(x) for 0 <= x < 2pi.

1/cos(x) = 2cos(x) + sin(x)/cos(x)

Multiply by cos(x):

1 = 2cos²(x) + sin(x)

Using cos²(x) = 1 - sin²(x):

1 = 2(1 - sin²(x)) + sin(x)

1 = 2 - 2sin²(x) + sin(x)

2sin²(x) - sin(x) - 1 = 0

(2sin(x) + 1)(sin(x) - 1) = 0

sin(x) = -1/2: x = 7pi/6, 11pi/6 sin(x) = 1: x = pi/2

But at x = pi/2, cos(x) = 0, so sec(x) and tan(x) are undefined. Reject.

Solutions: x = 7pi/6 and x = 11pi/6

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Proving Identities

Worked Example 6

Prove that sec(theta) - cos(theta) = sin(theta) x tan(theta).

LHS = 1/cos(theta) - cos(theta)

= (1 - cos²(theta))/cos(theta)

= sin²(theta)/cos(theta)

= sin(theta) x (sin(theta)/cos(theta))

= sin(theta) x tan(theta)

= RHS

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Exam Tips

Tip	Detail
Convert to sin/cos	When stuck, write everything in terms of sin and cos
Use identities	1 + tan² = sec² and 1 + cot² = cosec² are essential
Check for undefined points	Always check that solutions don't make the original equation undefined
Reject impossible values	If cosec(x) = 0.5, then sin(x) = 2, which is impossible — reject it
Graph sketching	Know the shapes and asymptote positions for all three reciprocal functions

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Practice Problems

1. Find the exact value of cot(pi/3). [Answer: 1/sqrt(3) = sqrt(3)/3]
2. Solve cosec²(x) - 2 = 0 for 0 <= x < 2pi. [Answer: pi/4, 3pi/4, 5pi/4, 7pi/4]
3. Prove that cosec²(theta) - cot²(theta) = 1.
4. Solve sec(x) + tan(x) = 3 for 0 <= x < 2pi. [Answer: x ≈ 1.176]
5. Sketch y = sec(x) for 0 <= x <= 2pi, marking all asymptotes and key points.

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Summary

• sec = 1/cos, cosec = 1/sin, cot = cos/sin = 1/tan.
• Key identities: 1 + tan²(theta) = sec²(theta) and 1 + cot²(theta) = cosec²(theta).
• Graphs have vertical asymptotes where the original function is zero.
• To solve equations with sec, cosec or cot, convert to sin/cos or use the identities.

A-Level Deep Dive: Reciprocal Trigonometric Functions

Spec mapping

Edexcel 9MA0 specification section 5 — Trigonometry, sub-strand on reciprocal functions covers the secant, cosecant and cotangent functions, their domains and ranges; their graphs (including asymptotes); use of $1 + \tan^2\theta = \sec^2\theta$1+tan2θ=sec2θ and $1 + \cot^2\theta = \csc^2\theta$1+cot2θ=csc2θ (refer to the official specification document for exact wording). Reciprocal trig is a Year 2 topic, examined principally on Paper 2 (Pure Mathematics) but synoptically active throughout. It connects to the Pythagorean identities (section 5, where $\sin^2\theta + \cos^2\theta = 1$sin2θ+cos2θ=1 is the parent identity from which the two derived forms are obtained), differentiation (section 9, where $\frac{d}{dx}\sec x = \sec x \tan x$dxd​secx=secxtanx is on the formula booklet but its derivation rests on the quotient rule applied to $1/\cos x$1/cosx), integration (section 10, with the standard results $\int \sec^2 x\,dx = \tan x + C$∫sec2xdx=tanx+C and $\int \csc^2 x\,dx = -\cot x + C$∫csc2xdx=−cotx+C used routinely), trigonometric equations with multiple solutions in a stated interval, and integration by substitution (section 10, where rewriting in terms of $\sec$sec or $\tan$tan is often the unlock). The Edexcel formula booklet lists $\sec^2\theta = 1 + \tan^2\theta$sec2θ=1+tan2θ and $\csc^2\theta = 1 + \cot^2\theta$csc2θ=1+cot2θ, but candidates are still expected to derive them on demand.

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Worked example with full mark scheme

Question (8 marks): Solve the equation

$\sec^2\theta - 3\tan\theta - 5 = 0$sec2θ−3tanθ−5=0

for $\theta \in [0°, 360°]$θ∈[0°,360°], giving each solution to one decimal place.

Solution with mark scheme:

Step 1 — convert to a single trig function using the derived Pythagorean identity.

The presence of $\sec^2\theta$sec2θ alongside $\tan\theta$tanθ is the signal: replace $\sec^2\theta$sec2θ using

$\sec^2\theta = 1 + \tan^2\theta$sec2θ=1+tan2θ

to obtain a single-function equation in $\tan\theta$tanθ.

M1 — using $\sec^2\theta = 1 + \tan^2\theta$sec2θ=1+tan2θ to convert to a quadratic in $\tan\theta$tanθ. This is the routine examiner-rewarded phrasing: state the substitution explicitly. Candidates who try to convert to $\sin$sin and $\cos$cos in fractional form invariably tangle themselves and lose this M1.

Substituting:

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

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

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

Step 2 — factorise the quadratic.

$(\tan\theta - 4)(\tan\theta + 1) = 0$(tanθ−4)(tanθ+1)=0

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

M1 — factorising or using the quadratic formula on the equation in $\tan\theta$tanθ.

A1 — both correct values of $\tan\theta$tanθ.

Step 3 — solve $\tan\theta = 4$tanθ=4 in $[0°, 360°]$[0°,360°].

The principal value is $\theta = \arctan 4 \approx 75.96°$θ=arctan4≈75.96°. The tangent function has period $180°$180°, so the second solution in the interval is

$\theta = 75.96° + 180° = 255.96°$θ=75.96°+180°=255.96°

To one decimal place: $\theta = 76.0°$θ=76.0° and $\theta = 256.0°$θ=256.0°.

M1 — correct method for finding all solutions of $\tan\theta = k$tanθ=k in the stated interval (principal value plus $180°$180°).

A1 — both correct values for the $\tan\theta = 4$tanθ=4 branch.

Step 4 — solve $\tan\theta = -1$tanθ=−1 in $[0°, 360°]$[0°,360°].

Principal value (for negative tangent): $\theta = 180° - 45° = 135°$θ=180°−45°=135°, and adding the period $180°$180° gives $\theta = 315°$θ=315°.

A1 — both correct values for the $\tan\theta = -1$tanθ=−1 branch.

Step 5 — present the full solution set.

$\theta \in \{76.0°, 135°, 256.0°, 315°\}$θ∈{76.0°,135°,256.0°,315°}

A1 — complete, correctly rounded solution set with no extraneous values and no missing solutions.

Total: 8 marks (M3 A5, split as shown).

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): (a) Show that the equation

$2\csc^2\theta - 3\cot\theta = 5$2csc2θ−3cotθ=5

can be written in the form $2\cot^2\theta - 3\cot\theta - 3 = 0$2cot2θ−3cotθ−3=0. (2)

(b) Hence solve, for $\theta \in [0°, 180°]$θ∈[0°,180°],

$2\csc^2\theta - 3\cot\theta = 5$2csc2θ−3cotθ=5

giving each solution to one decimal place. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — using $\csc^2\theta = 1 + \cot^2\theta$csc2θ=1+cot2θ to convert to a single trig function. The substitution must be quoted, not assumed.
• A1 (AO1.1b) — algebraic rearrangement to the printed form $2\cot^2\theta - 3\cot\theta - 3 = 0$2cot2θ−3cotθ−3=0.

(b)

• M1 (AO1.1b) — applying the quadratic formula (the quadratic in $\cot\theta$cotθ does not factorise over rationals, so the formula is essential): $\cot\theta = \dfrac{3 \pm \sqrt{9 + 24}}{4} = \dfrac{3 \pm \sqrt{33}}{4}$cotθ=43±9+24​​=43±33​​.
• M1 (AO1.1b) — converting each value of $\cot\theta$cotθ to $\tan\theta = 1/\cot\theta$tanθ=1/cotθ and finding $\theta = \arctan(1/\cot\theta)$θ=arctan(1/cotθ) in the stated interval.
• A1 (AO1.1b) — first correct solution to one decimal place.
• A1 (AO2.1) — second correct solution to one decimal place, with a clear statement that solutions outside $[0°, 180°]$[0°,180°] have been rejected.

Total: 6 marks split AO1 = 5, AO2 = 1. This is an AO1-dominated question — the AO2 mark is reserved for the candidate who explicitly rejects out-of-interval solutions rather than silently dropping them. Edexcel rewards interval-discipline.

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Synoptic links

Connects to:

1. Section 5 — Pythagorean derived identities: the identities $1 + \tan^2\theta = \sec^2\theta$1+tan2θ=sec2θ and $1 + \cot^2\theta = \csc^2\theta$1+cot2θ=csc2θ are derived by dividing $\sin^2\theta + \cos^2\theta = 1$sin2θ+cos2θ=1 through by $\cos^2\theta$cos2θ and $\sin^2\theta$sin2θ respectively. Knowing the derivation lets you reconstruct the identities under exam stress if the formula booklet is misremembered.

2. Section 9 — Differentiation of $\sec x$secx: $\dfrac{d}{dx}\sec x = \sec x \tan x$dxd​secx=secxtanx follows from the quotient rule applied to $1/\cos x$1/cosx:

   $\dfrac{d}{dx}\left(\dfrac{1}{\cos x}\right) = \dfrac{0 \cdot \cos x - 1 \cdot (-\sin x)}{\cos^2 x} = \dfrac{\sin x}{\cos^2 x} = \sec x \tan x.$dxd​(cosx1​)=cos2x0⋅cosx−1⋅(−sinx)​=cos2xsinx​=secxtanx.

3. Section 10 — Integration of $\sec^2 x$sec2x and $\csc^2 x$csc2x: the standard results $\int \sec^2 x\,dx = \tan x + C$∫sec2xdx=tanx+C and $\int \csc^2 x\,dx = -\cot x + C$∫csc2xdx=−cotx+C are reverse-derivative facts. They appear constantly in integration by substitution — for example, $\int \tan^2 x\,dx$∫tan2xdx is most efficiently handled by writing $\tan^2 x = \sec^2 x - 1$tan2x=sec2x−1 and integrating term by term.

4. Section 5 — Trigonometric equations with multiple solutions: every reciprocal-trig equation question is, after substitution, a multi-solution interval-finding problem. Robust technique here (principal value, period, reflection symmetries) is the same machinery used for $\sin\theta = k$sinθ=k and $\cos\theta = k$cosθ=k.

5. Section 5 — Trigonometric identities (compound and double angle): the compound-angle formula $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$tan(A+B)=1−tanAtanBtanA+tanB​ is sometimes expressed using $\cot$cot via reciprocation. Identifying when reciprocal form is cleaner is an A* reasoning skill.

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Mark-scheme literacy

Reciprocal-trig questions on 9MA0 split AO marks toward AO1, with a meaningful AO2 reasoning component:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Quoting and using the derived Pythagorean identity, factorising the resulting quadratic, finding all solutions in the stated interval, presenting answers to the requested precision
AO2 (reasoning / interpretation)	20–30%	Choosing whether to convert to $\sin/\cos$sin/cos or apply the derived identity, justifying rejection of out-of-interval solutions, recognising when squaring may introduce extraneous roots
AO3 (problem-solving)	0–10%	Multi-step questions where the reciprocal-trig manipulation is one of several sub-steps in a longer Paper 2 problem

Examiner-rewarded phrasing: "using $\sec^2\theta = 1 + \tan^2\theta$sec2θ=1+tan2θ to convert to a quadratic in $\tan\theta$tanθ"; "the principal value is $\arctan(\cdot)$arctan(⋅), and adding the period $180°$180° gives the second solution in $[0°, 360°]$[0°,360°]"; "rejecting $\theta = -45°$θ=−45° since it lies outside the stated interval". Phrases that lose marks: "solving $\sec\theta = k$secθ=k by writing $\theta = \sec^{-1}k$θ=sec−1k" without acknowledging that $\sec^{-1}$sec−1 is not in the standard A-Level toolkit (convert to $\cos\theta = 1/k$cosθ=1/k instead); leaving answers in unrounded form when one decimal place is requested; quoting the period of $\sec\theta$secθ as $180°$180° (it is $360°$360°, inherited from cosine).

A specific Edexcel pattern to watch: questions that look solvable purely in $\sin/\cos$sin/cos but reduce more cleanly via the derived identity — the AO2 mark for "method choice" goes to candidates who pick the shorter route. The signal is the presence of $\sec^2\theta$sec2θ or $\csc^2\theta$csc2θ alongside $\tan\theta$tanθ or $\cot\theta$cotθ; reach for the derived identity immediately.

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Grade-band model answers

3-mark question

Question: State the exact value of $\sec(\pi/3)$sec(π/3), justifying your answer.

Grade C response (~210 words):

By definition, $\sec\theta = \dfrac{1}{\cos\theta}$secθ=cosθ1​.

We know $\cos(\pi/3) = \dfrac{1}{2}$cos(π/3)=21​ from the standard exact values.

So $\sec(\pi/3) = \dfrac{1}{1/2} = 2$sec(π/3)=1/21​=2.

The exact value is $\sec(\pi/3) = 2$sec(π/3)=2.