Home / A-Level Maths 9709 / Trigonometry / Secant, cosecant and cotangent
Cambridge International AS & A Level · Topic 3.3
A-Level Maths 9709 / Topic 3.3 / AS & A Level

Three new functions, three old friends.

Papers 2 and 3. Secant, cosecant and cotangent are the reciprocals of cosine, sine and tangent. Learn their graphs, then use \(\sec^2\theta \equiv 1 + \tan^2\theta\) and \(\operatorname{cosec}^2\theta \equiv 1 + \cot^2\theta\) to turn awkward equations into quadratics.

Topic 3.3All trigonometry lessons →
The key idea

\(\sec\theta = \dfrac{1}{\cos\theta}\), \(\operatorname{cosec}\theta = \dfrac{1}{\sin\theta}\), \(\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}\). Dividing \(\sin^2\theta + \cos^2\theta \equiv 1\) by \(\cos^2\theta\) or \(\sin^2\theta\) gives \(1 + \tan^2\theta \equiv \sec^2\theta\) and \(1 + \cot^2\theta \equiv \operatorname{cosec}^2\theta\).

x y 90°180°270°360° 1−1 y = sec x y = cos x
Fig. 1: y = sec x (blue) is 1 ÷ cos x (dashed). Where cos x = ±1 they touch; where cos x = 0 sec x has vertical asymptotes, and sec x never lies between −1 and 1
Section 01

The reciprocal functions.

Each new function is one over an old one. Their graphs follow from the old graphs: where the old function is small, the new one is huge, and where the old function is zero, the new one has a vertical asymptote.

FunctionDefinitionUndefined whenRangePeriod
\(\sec x\)\(\dfrac{1}{\cos x}\)\(\cos x = 0\): \(x = 90^\circ, 270^\circ, \ldots\)\(\sec x \le -1\) or \(\sec x \ge 1\)\(360^\circ\)
\(\operatorname{cosec} x\)\(\dfrac{1}{\sin x}\)\(\sin x = 0\): \(x = 0^\circ, 180^\circ, \ldots\)\(\operatorname{cosec} x \le -1\) or \(\operatorname{cosec} x \ge 1\)\(360^\circ\)
\(\cot x\)\(\dfrac{\cos x}{\sin x}\)\(\sin x = 0\): \(x = 0^\circ, 180^\circ, \ldots\)all real numbers\(180^\circ\)
Worked example Exact values
Find the exact values of (a) \(\sec 60^\circ\) (b) \(\operatorname{cosec} 225^\circ\) (c) \(\cot \dfrac{5\pi}{6}\).
  1. (a) \(\cos 60^\circ = \dfrac{1}{2}\), so \(\sec 60^\circ = 2\).
  2. (b) \(225^\circ\) is in the third quadrant, so \(\sin 225^\circ = -\sin 45^\circ = -\dfrac{1}{\sqrt{2}}\).
  3. \(\operatorname{cosec} 225^\circ = -\sqrt{2}\).
  4. (c) \(\cos\dfrac{5\pi}{6} = -\dfrac{\sqrt{3}}{2}\) and \(\sin\dfrac{5\pi}{6} = \dfrac{1}{2}\), so \(\cot\dfrac{5\pi}{6} = \dfrac{-\sqrt{3}/2}{1/2} = -\sqrt{3}\).
Answer: (a) \(2\)   (b) \(-\sqrt{2}\)   (c) \(-\sqrt{3}\)
Worked example A simple equation
Solve \(\sec x = -2\) for \(0^\circ \le x \le 360^\circ\).
  1. Turn it upside down: \(\cos x = -\dfrac{1}{2}\).
  2. The reference angle is \(60^\circ\); cosine is negative in the second and third quadrants.
  3. \(x = 180^\circ - 60^\circ = 120^\circ\) and \(x = 180^\circ + 60^\circ = 240^\circ\).
Answer: \(x = 120^\circ,\ 240^\circ\)
Examiner trap

\(\sec x\) is NOT \(\cos^{-1} x\). The index \(-1\) on a function name means the inverse function; \(\sec x = (\cos x)^{-1}\) is the reciprocal. On the calculator, \(\sec 50^\circ\) is \(1 \div \cos 50^\circ = 1.56\), not \(\cos^{-1} 50\) (which gives an error).

Section 02

Two Pythagorean identities.

Start from \(\sin^2\theta + \cos^2\theta \equiv 1\). Divide every term by \(\cos^2\theta\) to get the first identity, and by \(\sin^2\theta\) to get the second. Their job in an equation is to make every term the same function, so the equation becomes a quadratic.

Worked example A quadratic in tan
Solve \(2\sec^2\theta + \tan\theta = 8\) for \(0^\circ \le \theta \le 360^\circ\).
  1. Replace \(\sec^2\theta\) by \(1 + \tan^2\theta\): \(2 + 2\tan^2\theta + \tan\theta = 8\).
  2. Rearrange: \(2\tan^2\theta + \tan\theta - 6 = 0\).
  3. Factorise: \((2\tan\theta - 3)(\tan\theta + 2) = 0\), so \(\tan\theta = 1.5\) or \(\tan\theta = -2\).
  4. \(\tan\theta = 1.5\): \(\theta = 56.3^\circ\) or \(56.3^\circ + 180^\circ = 236.3^\circ\).
  5. \(\tan\theta = -2\): the calculator gives \(-63.4^\circ\); add \(180^\circ\) twice: \(116.6^\circ,\ 296.6^\circ\).
Answer: \(\theta = 56.3^\circ,\ 116.6^\circ,\ 236.3^\circ,\ 296.6^\circ\)
Worked example Mixing tan and cot
Solve \(\tan\theta + \cot\theta = 3\) for \(0^\circ < \theta < 180^\circ\).
  1. Write \(\cot\theta = \dfrac{1}{\tan\theta}\) and let \(t = \tan\theta\): \(t + \dfrac{1}{t} = 3\).
  2. Multiply by \(t\): \(t^2 - 3t + 1 = 0\).
  3. Quadratic formula: \(t = \dfrac{3 \pm \sqrt{5}}{2}\), so \(t = 2.618\) or \(t = 0.382\).
  4. Both are positive, so each gives one angle in \(0^\circ < \theta < 90^\circ\) and none in the second quadrant (tan is negative there).
  5. \(\theta = \tan^{-1} 2.618 = 69.1^\circ\) and \(\theta = \tan^{-1} 0.382 = 20.9^\circ\).
Answer: \(\theta = 20.9^\circ,\ 69.1^\circ\)
Your turn
Solve \(3\operatorname{cosec}^2\theta = 5 + \cot\theta\) for \(0^\circ \le \theta \le 360^\circ\).
Examiner trap

The identity is \(\sec^2\theta = 1 + \tan^2\theta\), not \(\sec^2\theta = 1 - \tan^2\theta\) or \(\tan^2\theta = 1 + \sec^2\theta\). If you forget, re-derive it in one line by dividing \(\sin^2\theta + \cos^2\theta = 1\) by \(\cos^2\theta\).

Stage 1 · Learn

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What is the exact value of \(\sec 60^\circ\)?

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Which value can \(\operatorname{cosec} x\) never take?

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Which identity is correct?

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What is \(\cot 135^\circ\)?

Section 03

Proving identities.

To prove an identity, start from one side (usually the more complicated one) and work until it becomes the other side. Converting everything to \(\sin\) and \(\cos\) is the safe default; combining fractions over a common denominator does the rest.

Worked example Prove an identity
Prove that \(\sec x - \cos x \equiv \sin x \tan x\).
  1. Start with the left-hand side and write \(\sec x = \dfrac{1}{\cos x}\): LHS \(= \dfrac{1}{\cos x} - \cos x\).
  2. Common denominator: \(= \dfrac{1 - \cos^2 x}{\cos x}\).
  3. Use \(1 - \cos^2 x = \sin^2 x\): \(= \dfrac{\sin^2 x}{\cos x}\).
  4. Split: \(= \sin x \times \dfrac{\sin x}{\cos x} = \sin x \tan x\) = RHS.
Answer: LHS \(= \dfrac{\sin^2 x}{\cos x} = \sin x \tan x\) = RHS, as required.
Worked example Prove an identity with squares
Prove that \(\operatorname{cosec}^2 x + \sec^2 x \equiv \operatorname{cosec}^2 x \sec^2 x\).
  1. LHS \(= \dfrac{1}{\sin^2 x} + \dfrac{1}{\cos^2 x}\).
  2. Common denominator: \(= \dfrac{\cos^2 x + \sin^2 x}{\sin^2 x \cos^2 x}\).
  3. The numerator is 1: \(= \dfrac{1}{\sin^2 x \cos^2 x} = \dfrac{1}{\sin^2 x} \times \dfrac{1}{\cos^2 x}\) = RHS.
Answer: LHS \(= \dfrac{1}{\sin^2 x\cos^2 x} = \operatorname{cosec}^2 x \sec^2 x\) = RHS.
Examiner trap

Never work on both sides at once, and never cross-multiply an identity you are asked to prove: that assumes the result. Write LHS \(= \ldots = \) RHS as one chain.

Stage 2 · Exam

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Solve \(2\tan^2\theta + \sec\theta = 1\) for \(0^\circ \le \theta \le 360^\circ\).

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Given that \(\cos x = -\dfrac{3}{5}\) and \(x\) is obtuse, find the exact value of \(\operatorname{cosec} x + \cot x\).

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Given that \(\tan\theta = k\), where \(k \ne 0\), express \(\operatorname{cosec}^2\theta\) in terms of \(k\).

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Solve \(3\cot^2\theta - 5\operatorname{cosec}\theta + 1 = 0\) for \(0^\circ < \theta < 360^\circ\).

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Six original Cambridge-style questions
Exact values, proving identities, and equations that become quadratics using \(\sec^2\theta = 1 + \tan^2\theta\) and \(\operatorname{cosec}^2\theta = 1 + \cot^2\theta\).