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Learn7357Paper 3 · Pure and StatisticsPure content (sections A–J)Reciprocal functions and their identities
Pure content (sections A–J) · Subtopic 41 of 66

Reciprocal Functions: sec, cosec and cot

Three new functions that are just 11 over the old ones: their values, their graphs with asymptotes, the two Pythagorean identities they satisfy, and how to use them to solve equations.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 40 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Define sec⁡θ=1cos⁡θ\sec\theta=\dfrac{1}{\cos\theta}, cosec⁡θ=1sin⁡θ\operatorname{cosec}\theta=\dfrac{1}{\sin\theta} and cot⁡θ=1tan⁡θ\cot\theta=\dfrac{1}{\tan\theta}.
  • Sketch their graphs, with asymptotes where the original function is zero.
  • Derive and use sec⁡2θ≡1+tan⁡2θ\sec^2\theta\equiv1+\tan^2\theta and cosec⁡2θ≡1+cot⁡2θ\operatorname{cosec}^2\theta\equiv1+\cot^2\theta.
  • Solve equations by turning them into a quadratic in one function.
Pause and think

Divide every term of sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 by cos⁡2θ\cos^2\theta. What do you get?

Section 2 of 14

Core idea

Start hereThree more names for old friends

Surveyors and navigators once worked so often with 1cos⁡θ\frac1{\cos\theta} that it got its own name, the secant. Its cousins cosecant and cotangent followed. They're not new ideas, just reciprocals, but they let identities and calculus answers be written neatly.

The big idea

sec⁡θ=1cos⁡θ\sec\theta=\frac1{\cos\theta}, cosec⁡θ=1sin⁡θ\operatorname{cosec}\theta=\frac1{\sin\theta}, cot⁡θ=1tan⁡θ=cos⁡θsin⁡θ\cot\theta=\frac1{\tan\theta}=\frac{\cos\theta}{\sin\theta}. Dividing sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 by cos⁡2θ\cos^2\theta or sin⁡2θ\sin^2\theta gives 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta and 1+cot⁡2θ=cosec⁡2θ1+\cot^2\theta=\operatorname{cosec}^2\theta.

1. Definitions and graphs

sec⁡θ=1cos⁡θcosec⁡θ=1sin⁡θcot⁡θ=cos⁡θsin⁡θ\sec\theta=\frac1{\cos\theta}\qquad\operatorname{cosec}\theta=\frac1{\sin\theta}\qquad\cot\theta=\frac{\cos\theta}{\sin\theta}
where
undefined
whenever the denominator is 00: sec⁡\sec at cos⁡θ=0\cos\theta=0 (90°, 270°), cosec⁡\operatorname{cosec} and cot⁡\cot at sin⁡θ=0\sin\theta=0 (0°, 180°, 360°).
y = cos x and y = sec x
50100150200250300350−4−3−2−112340x (degrees)ycos xsec x
Where cos⁡x=1\cos x=1, sec⁡x=1\sec x=1 too. Where cos⁡x\cos x is small, sec⁡x\sec x is huge, and where cos⁡x=0\cos x=0 (90∘90^\circ, 270∘270^\circ) sec⁡x\sec x has asymptotes. sec⁡x\sec x is never between −1-1 and 11.
  • Where cos⁡x=±1\cos x=\pm1, sec⁡x=±1\sec x=\pm1; as cos⁡x→0\cos x\to0, ∣sec⁡x∣→∞|\sec x|\to\infty (asymptotes).
  • ∣sec⁡x∣≥1|\sec x|\ge1 and ∣cosec⁡x∣≥1|\operatorname{cosec}x|\ge1 always.
  • cot⁡x\cot x: period 180∘180^\circ, zero at 90∘90^\circ.

2. The new Pythagorean identities

Deriving 1+tan⁡2θ≡sec⁡2θ1+\tan^2\theta\equiv\sec^2\theta
  1. sin⁡2θ+cos⁡2θ≡1\sin^2\theta+\cos^2\theta\equiv1
    Start.
  2. sin⁡2θcos⁡2θ+cos⁡2θcos⁡2θ≡1cos⁡2θ\frac{\sin^2\theta}{\cos^2\theta}+\frac{\cos^2\theta}{\cos^2\theta}\equiv\frac1{\cos^2\theta}
    Divide every term by cos⁡2θ\cos^2\theta (where cos⁡θ≠0\cos\theta\neq0).
  3. tan⁡2θ+1≡sec⁡2θ\tan^2\theta+1\equiv\sec^2\theta
    Recognise each fraction.
Dividing by sin⁡2θ\sin^2\theta instead gives 1+cot⁡2θ≡cosec⁡2θ1+\cot^2\theta\equiv\operatorname{cosec}^2\theta.
Worked example2tan⁡2θ+sec⁡θ=12\tan^2\theta+\sec\theta=1, 0≤θ≤360∘0\le\theta\le360^\circ

Solve.

Goal
All θ\theta.
Strategy
Replace tan⁡2\tan^2 to get a quadratic in sec⁡θ\sec\theta.
  1. 2(sec⁡2θ−1)+sec⁡θ=12(\sec^2\theta-1)+\sec\theta=1
    tan⁡2=sec⁡2−1\tan^2=\sec^2-1.
  2. 2sec⁡2θ+sec⁡θ−3=02\sec^2\theta+\sec\theta-3=0
    Rearrange.
  3. (2sec⁡θ+3)(sec⁡θ−1)=0(2\sec\theta+3)(\sec\theta-1)=0
    Factorise.
  4. cos⁡θ=−23 or cos⁡θ=1\cos\theta=-\tfrac23\ \text{or}\ \cos\theta=1
    Take reciprocals.
  5. θ=131.8∘,228.2∘;  0∘,360∘\theta=131.8^\circ,228.2^\circ;\ \ 0^\circ,360^\circ
    Solve each.
Answer0∘,131.8∘,228.2∘,360∘0^\circ,131.8^\circ,228.2^\circ,360^\circ.
✓ Checkθ=0\theta=0: 0+1=10+1=1 ✓.
Worked exampleProve cosec⁡θcot⁡θ+tan⁡θ≡cos⁡θ\dfrac{\operatorname{cosec}\theta}{\cot\theta+\tan\theta}\equiv\cos\theta

Prove.

Goal
LHS → RHS.
Strategy
Write everything in sin and cos.
  1. 1sin⁡θcos⁡θsin⁡θ+sin⁡θcos⁡θ\frac{\frac1{\sin\theta}}{\frac{\cos\theta}{\sin\theta}+\frac{\sin\theta}{\cos\theta}}
    Definitions.
  2. =1sin⁡θcos⁡2θ+sin⁡2θsin⁡θcos⁡θ=1sin⁡θ1sin⁡θcos⁡θ=\frac{\frac1{\sin\theta}}{\frac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}}=\frac{\frac1{\sin\theta}}{\frac1{\sin\theta\cos\theta}}
    Common denominator; Pythagoras.
  3. =1sin⁡θ×sin⁡θcos⁡θ=cos⁡θ=\frac1{\sin\theta}\times\sin\theta\cos\theta=\cos\theta
    Divide by a fraction = multiply by its reciprocal.
AnswerProved ∎.
✓ Checkθ=60∘\theta=60^\circ: 2/31/3+3=2/34/3=12=cos⁡60∘\frac{2/\sqrt3}{1/\sqrt3+\sqrt3}=\frac{2/\sqrt3}{4/\sqrt3}=\frac12=\cos60^\circ ✓.
Common misconceptionsec⁡θ\sec\theta is not cos⁡−1θ\cos^{-1}\theta sec⁡θ=1cos⁡θ\sec\theta=\frac1{\cos\theta} (a reciprocal); cos⁡−1\cos^{-1} is the inverse function.
Pause and think

tan⁡θ=34\tan\theta=\frac34, θ\theta acute. Find sec⁡θ\sec\theta and cosec⁡θ\operatorname{cosec}\theta.

Summary

  • Reciprocal definitions; asymptotes where denominators vanish.
  • 1+tan⁡2=sec⁡21+\tan^2=\sec^2; 1+cot⁡2=cosec⁡21+\cot^2=\operatorname{cosec}^2.
Key words
Secant
1/cos⁡1/\cos.
Cosecant
1/sin⁡1/\sin.
Cotangent
1/tan⁡1/\tan.
HistoryCutting and touching. "Tangent" comes from the Latin for touching and "secant" from the Latin secare, to cut. The Danish mathematician Thomas Fincke introduced both names in 1583. On the unit circle they are real lengths: tan⁡θ\tan\theta is a piece of the tangent line, and sec⁡θ\sec\theta is a piece of the line that cuts through the circle.
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