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How do we use the Pythagorean and reciprocal identities to simplify expressions and prove trigonometric statements?

State and use the Pythagorean, reciprocal and quotient identities to simplify expressions and prove trigonometric identities

A focused answer to the O-Level A-Maths outcome on trigonometric identities. The Pythagorean, reciprocal and quotient identities, and a reliable strategy for proving identities by working one side.

Generated by Claude Opus 4.89 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

SEAB wants you to know the basic trigonometric identities, to use them to simplify expressions into a single ratio, and to prove given identities by manipulating one side until it matches the other. Proving identities rewards a clear, one-direction argument rather than treating the statement as an equation to solve.

The answer

The reciprocal identities

The three reciprocal ratios are defined as:

sec⁡θ=1cos⁡θ,csc⁡θ=1sin⁡θ,cot⁡θ=1tan⁡θ=cos⁡θsin⁡θ.\sec\theta = \frac{1}{\cos\theta}, \qquad \csc\theta = \frac{1}{\sin\theta}, \qquad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}.

The quotient identity

Tangent is the ratio of sine to cosine:

tan⁡θ=sin⁡θcos⁡θ.\tan\theta = \frac{\sin\theta}{\cos\theta}.

The Pythagorean identities

From x2+y2=1x^2 + y^2 = 1 on the unit circle comes the master identity, and dividing it through gives two more:

sin⁡2θ+cos⁡2θ=1,\sin^2\theta + \cos^2\theta = 1,

1+tan⁡2θ=sec⁡2θ,1+cot⁡2θ=csc⁡2θ.1 + \tan^2\theta = \sec^2\theta, \qquad 1 + \cot^2\theta = \csc^2\theta.

These let you swap between sin⁡\sin, cos⁡\cos, tan⁡\tan and their reciprocals.

Strategy for proving an identity

Treat the two sides separately. Start from the more complicated side and transform it, step by justified step, until it becomes the other side. Useful moves: convert everything to sin⁡\sin and cos⁡\cos, find a common denominator, factorise, and apply a Pythagorean identity. Never move terms across the ≡\equiv sign as if solving an equation.

Examples in context

Example 1. Simplifying before solving. A messy equation such as sec⁡2θ−tan⁡θ=1\sec^2\theta - \tan\theta = 1 becomes a quadratic in tan⁡θ\tan\theta once sec⁡2θ=1+tan⁡2θ\sec^2\theta = 1 + \tan^2\theta is substituted, which is why identities are the first move in many equation problems.

Example 2. Verifying an integration result. Checking that a trigonometric integral has been done correctly often requires rewriting the answer with an identity to match a different but equivalent form, the same one-side manipulation used in proofs.

Try this

Q1. Simplify sin⁡θcot⁡θ\sin\theta\cot\theta. [2 marks]

  • Cue. sin⁡θ⋅cos⁡θsin⁡θ=cos⁡θ\sin\theta \cdot \dfrac{\cos\theta}{\sin\theta} = \cos\theta.

Q2. Show that cos⁡2θ−sin⁡2θ≡1−2sin⁡2θ\cos^2\theta - \sin^2\theta \equiv 1 - 2\sin^2\theta. [2 marks]

  • Cue. Replace cos⁡2θ\cos^2\theta with 1−sin⁡2θ1 - \sin^2\theta: (1−sin⁡2θ)−sin⁡2θ=1−2sin⁡2θ(1 - \sin^2\theta) - \sin^2\theta = 1 - 2\sin^2\theta.

Q3. Simplify 1−cos⁡2θcos⁡2θ\dfrac{1 - \cos^2\theta}{\cos^2\theta}. [2 marks]

  • Cue. sin⁡2θcos⁡2θ=tan⁡2θ\dfrac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksProve the identity 11+sin⁡θ+11−sin⁡θ≡2sec⁡2θ\dfrac{1}{1 + \sin\theta} + \dfrac{1}{1 - \sin\theta} \equiv 2\sec^2\theta.
Show worked answer →

Work on the left side. Add the fractions over the common denominator:

(1−sin⁡θ)+(1+sin⁡θ)(1+sin⁡θ)(1−sin⁡θ)=21−sin⁡2θ\dfrac{(1 - \sin\theta) + (1 + \sin\theta)}{(1 + \sin\theta)(1 - \sin\theta)} = \dfrac{2}{1 - \sin^2\theta}.

Use 1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta: 2cos⁡2θ=2sec⁡2θ\dfrac{2}{\cos^2\theta} = 2\sec^2\theta, which is the right side.

Markers reward combining over a common denominator, the difference of two squares, the Pythagorean identity, and the conversion to sec⁡2θ\sec^2\theta.

Original3 marksSimplify sin⁡θ1−cos⁡2θ\dfrac{\sin\theta}{1 - \cos^2\theta}, giving your answer as a single trigonometric ratio.
Show worked answer →

Use 1−cos⁡2θ=sin⁡2θ1 - \cos^2\theta = \sin^2\theta in the denominator: sin⁡θsin⁡2θ=1sin⁡θ\dfrac{\sin\theta}{\sin^2\theta} = \dfrac{1}{\sin\theta}.

So the expression simplifies to csc⁡θ\csc\theta (also written cosec⁡θ\operatorname{cosec}\theta).

Markers reward the Pythagorean substitution and the cancellation to csc⁡θ\csc\theta.

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