IGCSE · Practice questions · Trigonometric identities and proofs

LHS, step, step, RHS.

Six original Cambridge-style questions on trigonometric identities: using \(\sec^{2} = 1 + \tan^{2}\) for exact values, simplifying, model proofs laid out as \(\text{LHS} = \dots = \text{RHS}\), and showing an expression is constant.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Pick a side, then transform it.

Start from the more complicated side and write \(\text{LHS} = \dots\).
Convert \(\tan\), \(\cot\), \(\sec\), \(\operatorname{cosec}\) to \(\sin\) and \(\cos\) when you are unsure.
Combine fractions over a common denominator; expand brackets carefully.
Replace \(\sin^{2} + \cos^{2}\), \(1 - \sin^{2}\), \(1 - \cos^{2}\), \(\sec^{2} - 1\), \(\operatorname{cosec}^{2} - 1\).
Factorise (difference of two squares) or multiply by a conjugate, top and bottom.
Finish with \(\dots = \text{RHS}\); never work on both sides at once.

Recall

[2 marks]

Given that \(\cos\theta = -\dfrac{2}{5}\) and \(90^{\circ} < \theta < 180^{\circ}\), find the exact value of \(\tan\theta\).

\(\sec\theta = -\dfrac{5}{2}\), so \(\tan^{2}\theta = \sec^{2}\theta - 1 = \dfrac{25}{4} - 1 = \dfrac{21}{4}\). ✓
In the second quadrant \(\tan\theta < 0\), so \(\tan\theta = -\dfrac{\sqrt{21}}{2}\). ✓

Calculation

[3 marks]

Show that \(\dfrac{\operatorname{cosec}x}{\cot x + \tan x}\) simplifies to a single trigonometric function, and identify it.

\[\cot x + \tan x = \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} = \frac{\cos^{2}x + \sin^{2}x}{\sin x\cos x} = \frac{1}{\sin x\cos x}\] ✓
\[\frac{\operatorname{cosec}x}{\cot x + \tan x} = \frac{1}{\sin x} \times \sin x\cos x\] ✓
\(= \cos x\). ✓

Proof

[4 marks]

Prove that \(\dfrac{\sin x}{\operatorname{cosec}x - \cot x} \equiv 1 + \cos x\).

The denominator: \(\operatorname{cosec}x - \cot x = \dfrac{1}{\sin x} - \dfrac{\cos x}{\sin x} = \dfrac{1 - \cos x}{\sin x}\). ✓
\[\text{LHS} = \sin x \div \frac{1 - \cos x}{\sin x} = \frac{\sin^{2}x}{1 - \cos x}\] ✓
\[= \frac{1 - \cos^{2}x}{1 - \cos x} = \frac{(1 - \cos x)(1 + \cos x)}{1 - \cos x}\] ✓
\(= 1 + \cos x = \text{RHS}\). ✓

Proof

[5 marks]

Prove that \(\dfrac{\tan\theta}{1 + \sec\theta} + \dfrac{1 + \sec\theta}{\tan\theta} \equiv 2\operatorname{cosec}\theta\).

Combine over the common denominator: \[\text{LHS} = \frac{\tan^{2}\theta + (1 + \sec\theta)^{2}}{\tan\theta(1 + \sec\theta)}\] ✓
Expand and use \(\tan^{2}\theta = \sec^{2}\theta - 1\): the top is \(\sec^{2}\theta - 1 + 1 + 2\sec\theta + \sec^{2}\theta = 2\sec^{2}\theta + 2\sec\theta\). ✓
Factorise the top: \[= \frac{2\sec\theta(\sec\theta + 1)}{\tan\theta(1 + \sec\theta)} = \frac{2\sec\theta}{\tan\theta}\] ✓
\[= 2 \times \frac{1}{\cos\theta} \times \frac{\cos\theta}{\sin\theta} = \frac{2}{\sin\theta}\] ✓
\(= 2\operatorname{cosec}\theta = \text{RHS}\). ✓

Proof

[4 marks]

Prove that \(\dfrac{1 - \sin x}{1 + \sin x} \equiv (\sec x - \tan x)^{2}\).

Start from the RHS, which has more functions: \[\text{RHS} = \left(\frac{1}{\cos x} - \frac{\sin x}{\cos x}\right)^{2} = \frac{(1 - \sin x)^{2}}{\cos^{2}x}\] ✓
Use \(\cos^{2}x = 1 - \sin^{2}x = (1 - \sin x)(1 + \sin x)\): \[= \frac{(1 - \sin x)^{2}}{(1 - \sin x)(1 + \sin x)}\] ✓
Cancel one factor \(1 - \sin x\): \(= \dfrac{1 - \sin x}{1 + \sin x} = \text{LHS}\). ✓

Analysis

[6 marks]

(a) Show that \(\sin^{4}x + \sin^{2}x\cos^{2}x \equiv \sin^{2}x\).

(b) Hence find the value of \(\sin^{4}x + \sin^{2}x\cos^{2}x + \cos^{2}x\), explaining why it does not depend on \(x\).

(a) \(\text{LHS} = \sin^{2}x(\sin^{2}x + \cos^{2}x) = \sin^{2}x \times 1 = \sin^{2}x = \text{RHS}\). ✓
(b) By (a), the expression is \(\sin^{2}x + \cos^{2}x\). ✓
\(\sin^{2}x + \cos^{2}x = 1\) for every \(x\), so the value is \(1\), whatever \(x\) is. ✓
(c) Difference of two squares: \(\text{LHS} = (\sec^{2}x - \tan^{2}x)(\sec^{2}x + \tan^{2}x)\). ✓
\(\sec^{2}x - \tan^{2}x = 1\), so \(\text{LHS} = 1 \times (\sec^{2}x + \tan^{2}x) = \sec^{2}x + \tan^{2}x = \text{RHS}\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.