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Cambridge IGCSE · Topic 10.6
IGCSE Add Maths 0606 / Topic 10 / IGCSE

One triangle, three identities.

Pythagoras in a triangle with hypotenuse 1 gives \(\sin^{2}A + \cos^{2}A = 1\). Divide it by \(\cos^{2}A\) or by \(\sin^{2}A\) and two more identities appear, linking \(\sec\) with \(\tan\) and \(\operatorname{cosec}\) with \(\cot\). With these three and a clear layout you can simplify expressions, prove identities and show that an expression never changes its value.

Topic 10All trigonometry lessons →
The key idea

\(\sin^{2}A + \cos^{2}A = 1\), \(\sec^{2}A = 1 + \tan^{2}A\) and \(\operatorname{cosec}^{2}A = 1 + \cot^{2}A\) hold for every angle where both sides are defined (all three are in the List of formulas). To prove an identity, start from the more complicated side and change it, one line at a time, until it becomes the other side: \(\text{LHS} = \dots = \text{RHS}\). Never treat the identity as an equation and work on both sides at once.

θ 1 cos θ sin θ Pythagoras: (sin θ)² + (cos θ)² = 1² sin²θ + cos²θ = 1 ÷ cos²θ:  tan²θ + 1 = sec²θ ÷ sin²θ:  1 + cot²θ = cosec²θ
Fig. 1: in a right-angled triangle with hypotenuse 1 the other sides are cos θ and sin θ, so Pythagoras gives sin²θ + cos²θ = 1. Dividing every term by cos²θ gives tan²θ + 1 = sec²θ; dividing by sin²θ gives 1 + cot²θ = cosec²θ. The identities hold for every angle, not only acute ones
Section 01

Three identities from one.

An identity is true for every value of the angle (where both sides are defined), not just for a few solutions. The first one is Pythagoras in Fig. 1. Divide every term of \(\sin^{2}A + \cos^{2}A = 1\) by \(\cos^{2}A\): since \(\dfrac{\sin A}{\cos A} = \tan A\) and \(\dfrac{1}{\cos A} = \sec A\), you get \(\tan^{2}A + 1 = \sec^{2}A\). Divide by \(\sin^{2}A\) instead and you get \(1 + \cot^{2}A = \operatorname{cosec}^{2}A\). Learn each one in all its rearranged forms; spotting them is most of the work.

IdentityUseful rearrangements
\(\sin^{2}A + \cos^{2}A = 1\)\(\sin^{2}A = 1 - \cos^{2}A\), \(\cos^{2}A = 1 - \sin^{2}A\)
\(\sec^{2}A = 1 + \tan^{2}A\)\(\tan^{2}A = \sec^{2}A - 1\), \(\sec^{2}A - \tan^{2}A = 1\)
\(\operatorname{cosec}^{2}A = 1 + \cot^{2}A\)\(\cot^{2}A = \operatorname{cosec}^{2}A - 1\), \(\operatorname{cosec}^{2}A - \cot^{2}A = 1\)
Worked example One ratio gives the others
Given that \(\tan\theta = \dfrac{5}{12}\) and \(180^{\circ} < \theta < 270^{\circ}\), find the exact values of \(\sec\theta\) and \(\cos\theta\).
  1. \(\sec^{2}\theta = 1 + \tan^{2}\theta = 1 + \dfrac{25}{144} = \dfrac{169}{144}\), so \(\sec\theta = \pm\dfrac{13}{12}\).
  2. In the third quadrant \(\cos\theta < 0\), so \(\sec\theta < 0\) too.
  3. \(\sec\theta = -\dfrac{13}{12}\), and \(\cos\theta = \dfrac{1}{\sec\theta}\).
Answer: \(\sec\theta = -\frac{13}{12}\), \(\cos\theta = -\frac{12}{13}\)
Worked example Simplify with an identity
Simplify \(\dfrac{\operatorname{cosec}^{2}x - 1}{\operatorname{cosec}^{2}x}\).
  1. The top is \(\operatorname{cosec}^{2}x - 1 = \cot^{2}x\).
  2. Write in sine and cosine: \(\cot^{2}x \div \operatorname{cosec}^{2}x = \dfrac{\cos^{2}x}{\sin^{2}x} \times \sin^{2}x\).
  3. The \(\sin^{2}x\) cancels.
Answer: \(\cos^{2}x\)
Your turn
Simplify \((\sec^{2}\theta - 1)\cos^{2}\theta\).
Examiner trap

\(\sin^{2}A\) means \((\sin A)^{2}\), not \(\sin(A^{2})\). The identity squares the ratios: it does NOT say \(\sin A + \cos A = 1\) (at \(A = 45^{\circ}\) the sum is \(\sqrt{2}\)). And both terms must be of the SAME angle: \(\sin^{2}A + \cos^{2}B\) is not 1 in general.

Section 02

Proving an identity: one side at a time.

A proof shows that one side can be turned into the other. Write \(\text{LHS} = \dots\), change the expression one step at a time, and finish with \(\dots = \text{RHS}\). If you get stuck, these moves almost always work.

Worked example Write in sin and cos
Prove that \(\tan\theta + \cot\theta \equiv \sec\theta\operatorname{cosec}\theta\).
  1. \[\text{LHS} = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}\]
  2. Combine over the common denominator \(\sin\theta\cos\theta\): \[= \frac{\sin^{2}\theta + \cos^{2}\theta}{\sin\theta\cos\theta}\]
  3. The top is 1: \[= \frac{1}{\sin\theta\cos\theta} = \frac{1}{\cos\theta} \times \frac{1}{\sin\theta}\]
  4. \(= \sec\theta\operatorname{cosec}\theta = \text{RHS}\).
Answer: \(\text{LHS} = \text{RHS}\), so the identity is proved
Worked example Combine two fractions
Prove that \(\dfrac{\cos\theta}{1 - \sin\theta} - \dfrac{\cos\theta}{1 + \sin\theta} \equiv 2\tan\theta\).
  1. The common denominator is \((1 - \sin\theta)(1 + \sin\theta) = 1 - \sin^{2}\theta\).
  2. \[\text{LHS} = \frac{\cos\theta(1 + \sin\theta) - \cos\theta(1 - \sin\theta)}{1 - \sin^{2}\theta}\]
  3. Expand the top: \(\cos\theta + \sin\theta\cos\theta - \cos\theta + \sin\theta\cos\theta = 2\sin\theta\cos\theta\).
  4. Use \(1 - \sin^{2}\theta = \cos^{2}\theta\): \[= \frac{2\sin\theta\cos\theta}{\cos^{2}\theta} = \frac{2\sin\theta}{\cos\theta}\]
  5. \(= 2\tan\theta = \text{RHS}\).
Answer: \(\text{LHS} = \text{RHS}\), so the identity is proved
Your turn
Prove that \(\dfrac{1}{1 + \cos x} + \dfrac{1}{1 - \cos x} \equiv 2\operatorname{cosec}^{2}x\).
Examiner trap

Do not write the identity down as an equation and then cross-multiply, square or add to both sides until you reach something true like \(1 = 1\). That does not prove anything, and it scores no marks. Work on ONE side only, and never 'cancel' a term from a sum: \(\dfrac{\cos\theta}{1 - \sin\theta}\) is not \(\dfrac{\cos\theta}{1} - \dfrac{\cos\theta}{\sin\theta}\).

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Which identity do you get by dividing every term of \(\sin^{2}A + \cos^{2}A = 1\) by \(\cos^{2}A\)?

Quick check+10 XP

Simplify \(\dfrac{1 - \cos^{2}x}{\cos^{2}x}\).

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Given that \(\sec\theta = 3\), find \(\tan^{2}\theta\).

Quick check+10 XP

Which is a correct way to prove an identity?

Section 03

Factorise, use a conjugate, find a constant.

Two algebra tools crack the harder proofs. The difference of two squares factorises \(\sin^{4}x - \cos^{4}x\) as \((\sin^{2}x - \cos^{2}x)(\sin^{2}x + \cos^{2}x)\). The conjugate of \(1 + \sin x\) is \(1 - \sin x\), and \((1 - \sin x)(1 + \sin x) = 1 - \sin^{2}x = \cos^{2}x\); multiplying the top and bottom of a fraction by a conjugate turns its denominator into a single square. Finally, to show an expression has a constant value, simplify it until no trig function is left. In the next lesson you will use these identities to rewrite equations so that they contain only one trig function before you solve them.

Worked example Multiply by the conjugate
Prove that \(\dfrac{\cos x}{1 + \sin x} \equiv \sec x - \tan x\).
  1. Multiply the top and bottom of the LHS by \(1 - \sin x\): \[\text{LHS} = \frac{\cos x(1 - \sin x)}{(1 + \sin x)(1 - \sin x)}\]
  2. The bottom is \(1 - \sin^{2}x = \cos^{2}x\): \[= \frac{\cos x(1 - \sin x)}{\cos^{2}x} = \frac{1 - \sin x}{\cos x}\]
  3. Split the fraction: \(= \dfrac{1}{\cos x} - \dfrac{\sin x}{\cos x} = \sec x - \tan x = \text{RHS}\).
Answer: \(\text{LHS} = \text{RHS}\), so the identity is proved
Worked example A constant value
Show that \((3\sin x + 2\cos x)^{2} + (2\sin x - 3\cos x)^{2}\) has the same value for every \(x\), and state that value.
  1. First bracket: \(9\sin^{2}x + 12\sin x\cos x + 4\cos^{2}x\).
  2. Second bracket: \(4\sin^{2}x - 12\sin x\cos x + 9\cos^{2}x\).
  3. Add: the \(\sin x\cos x\) terms cancel, leaving \(13\sin^{2}x + 13\cos^{2}x\).
  4. \(= 13(\sin^{2}x + \cos^{2}x) = 13\), with no \(x\) left.
Answer: The expression always equals \(13\)
Your turn
Prove that \(\sin^{4}x - \cos^{4}x \equiv 1 - 2\cos^{2}x\).
Examiner trap

Squaring a bracket gives three terms: \((1 + \sin x)^{2} = 1 + 2\sin x + \sin^{2}x\), not \(1 + \sin^{2}x\). And when you multiply by a conjugate, multiply the top AND the bottom, or you have changed the value of the fraction.

Stage 2 · Exam

Exam-style questions

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Given that \(\cot x = 2\) and \(180^{\circ} < x < 270^{\circ}\), find the exact value of \(\sin x\).

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Simplify \((\sec x - \cos x)\cot x\).

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The expression \((2\cos x + \sin x)^{2} + (\cos x - 2\sin x)^{2}\) is constant. Find its value.

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To prove \(\dfrac{\sin x}{1 - \cos x} \equiv \dfrac{1 + \cos x}{\sin x}\) from the LHS, the best first step is to multiply

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Identities proved, one side at a time.

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Six original Cambridge-style questions
Exact values from one ratio, simplifying with identities, three proofs using conjugates, fractions and factorising, and a constant value.