Trigonometric Identities: tan θ and sin²θ + cos²θ

Introduction

Trigonometric identities are equations that hold true for all values of the variable (for which both sides are defined). The two identities covered here are the bedrock of A-Level trigonometry: virtually every manipulation, proof, or equation-solving task in the 9709 exam that goes beyond basic ratio work relies on one or both of them. Examiners regularly award method marks specifically for recognising which identity to apply and how to rearrange it — so fluency with these two results is non-negotiable.

Core Concept

The Quotient Identity

On the unit circle (or from the right-triangle definitions), we have:

\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}

Dividing gives:

\frac{\sin\theta}{\cos\theta} = \frac{\text{opposite}}{\text{adjacent}} = \tan\theta

This is true for all $\theta$ where $\cos\theta \neq 0$ (i.e. $\theta \neq 90°, 270°, \ldots$ ).

The Pythagorean Identity

On the unit circle, a point at angle $\theta$ has coordinates $(\cos\theta,\, \sin\theta)$ . Since it lies on the circle $x^2 + y^2 = 1$ :

\cos^2\theta + \sin^2\theta = 1

conventionally written as $\sin^2\theta + \cos^2\theta \equiv 1$ .

Two Rearrangements You Must Know

These rearrangements are used constantly and must be recalled instantly:

\sin^2\theta \equiv 1 - \cos^2\theta

\cos^2\theta \equiv 1 - \sin^2\theta

Key Formulae & Definitions

The two fundamental identities:

\frac{\sin\theta}{\cos\theta} \equiv \tan\theta \qquad (\cos\theta \neq 0)

\sin^2\theta + \cos^2\theta \equiv 1

Derived rearrangements:

\sin^2\theta \equiv 1 - \cos^2\theta

\cos^2\theta \equiv 1 - \sin^2\theta

Notation: The symbol $\equiv$ (identity) is preferred over $=$ when the relation holds for all valid $\theta$ . In the 9709 exam you may use either, but $\equiv$ signals mathematical precision.

Worked Examples

Example 1 — Proving an Identity

Prove that $\dfrac{1 - \cos^2\theta}{\cos^2\theta} \equiv \tan^2\theta$ .

Step 1: Work on the left-hand side (LHS) only.

\text{LHS} = \frac{1 - \cos^2\theta}{\cos^2\theta}

Step 2: Apply $\sin^2\theta + \cos^2\theta \equiv 1$ , so $1 - \cos^2\theta \equiv \sin^2\theta$ :

= \frac{\sin^2\theta}{\cos^2\theta}

Step 3: Write as a square of a fraction:

= \left(\frac{\sin\theta}{\cos\theta}\right)^2

Step 4: Apply $\dfrac{\sin\theta}{\cos\theta} \equiv \tan\theta$ :

= \tan^2\theta = \text{RHS} \qquad \blacksquare

Example 2 — Solving an Equation

Solve $2\sin^2\theta - \cos\theta - 1 = 0$ for $0° \leq \theta \leq 360°$ .

Step 1: The equation mixes $\sin^2\theta$ and $\cos\theta$ . Replace $\sin^2\theta$ using $\sin^2\theta \equiv 1 - \cos^2\theta$ :

2(1 - \cos^2\theta) - \cos\theta - 1 = 0

Step 2: Expand and collect:

2 - 2\cos^2\theta - \cos\theta - 1 = 0

-2\cos^2\theta - \cos\theta + 1 = 0

Step 3: Multiply through by $-1$ :

2\cos^2\theta + \cos\theta - 1 = 0

Step 4: Factorise as a quadratic in $\cos\theta$ . Let $c = \cos\theta$ :

2c^2 + c - 1 = (2c - 1)(c + 1) = 0

Step 5: Solve each factor:

\cos\theta = \tfrac{1}{2} \quad \Rightarrow \quad \theta = 60°,\; 300°

\cos\theta = -1 \quad \Rightarrow \quad \theta = 180°

Answer: $\theta = 60°,\; 180°,\; 300°$

Example 3 — Simplifying an Expression

Simplify $\sin^2\theta + \cos^2\theta + \tan^2\theta\cos^2\theta$ .

Step 1: Apply the Pythagorean identity to the first two terms:

\sin^2\theta + \cos^2\theta = 1

So the expression becomes $1 + \tan^2\theta\cos^2\theta$ .

Step 2: Replace $\tan^2\theta$ with $\dfrac{\sin^2\theta}{\cos^2\theta}$ :

1 + \frac{\sin^2\theta}{\cos^2\theta} \cdot \cos^2\theta = 1 + \sin^2\theta

Simplified result: $1 + \sin^2\theta$

Common Mistakes & Examiner Pitfalls

Mistake	Why it's wrong	Correct approach
Writing $\sin^2\theta = (\sin\theta)^2$ as $\sin\theta^2$	$\sin\theta^2$ means $\sin(\theta^2)$ , which is different	Always write $\sin^2\theta$ or $(\sin\theta)^2$
Dividing by $\sin\theta$ or $\cos\theta$ without checking they're non-zero	You may lose solutions (e.g. $\theta = 0°$ )	Factorise instead of dividing
Trying to prove an identity by working on both sides simultaneously	This is not valid proof structure	Work on one side (usually LHS) and reach the other
Using $\sin^2\theta + \cos^2\theta = 1$ to write $\sin\theta + \cos\theta = 1$	Taking the square root does not remove the squares like this	The identity involves squared terms only
Forgetting to apply the identity before solving — leaving a mixed-function equation	You cannot solve an equation with both $\sin^2\theta$ and $\cos\theta$ directly	Always reduce to a single trig function first
Missing solutions in a given interval	Stopping after the principal value	Always consider all quadrants in the given range

Practice Questions

Q1. Prove the identity: $\dfrac{\sin^2\theta}{1 - \cos\theta} \equiv 1 + \cos\theta$

<details><summary>Show answer</summary>

LHS $= \dfrac{\sin^2\theta}{1 - \cos\theta}$

Apply $\sin^2\theta \equiv 1 - \cos^2\theta$ :

= \frac{1 - \cos^2\theta}{1 - \cos\theta}

Factorise the numerator as a difference of two squares:

= \frac{(1 - \cos\theta)(1 + \cos\theta)}{1 - \cos\theta}

Cancel $(1 - \cos\theta)$ (valid since $\cos\theta \neq 1$ ):

= 1 + \cos\theta = \text{RHS} \qquad \blacksquare

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Q2. Solve $3\sin^2\theta = 2 - \cos\theta$ for $0° \leq \theta \leq 360°$ .

<details><summary>Show answer</summary>

Replace $\sin^2\theta$ with $1 - \cos^2\theta$ :

3(1 - \cos^2\theta) = 2 - \cos\theta

3 - 3\cos^2\theta = 2 - \cos\theta

3\cos^2\theta - \cos\theta - 1 = 0

Using the quadratic formula with $c = \cos\theta$ :

c = \frac{1 \pm \sqrt{1 + 12}}{6} = \frac{1 \pm \sqrt{13}}{6}

c = \frac{1 + \sqrt{13}}{6} \approx 0.7676 \quad \Rightarrow \quad \theta \approx 39.8°,\; 320.2°

c = \frac{1 - \sqrt{13}}{6} \approx -0.4343 \quad \Rightarrow \quad \theta \approx 115.7°,\; 244.3°

Answers (to 1 d.p.): $\theta \approx 39.8°,\; 115.7°,\; 244.3°,\; 320.2°$

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Q3. Simplify $\dfrac{\sin^2\theta + \sin\theta\cos\theta}{\cos^2\theta - \cos\theta\sin\theta}$ .

<details><summary>Show answer</summary>

Factor numerator and denominator:

\frac{\sin\theta(\sin\theta + \cos\theta)}{\cos\theta(\cos\theta - \sin\theta)}

This does not simplify to a standard identity directly, but we can write:

= \frac{\sin\theta}{\cos\theta} \cdot \frac{\sin\theta + \cos\theta}{\cos\theta - \sin\theta} = \tan\theta \cdot \frac{\sin\theta + \cos\theta}{\cos\theta - \sin\theta}

Dividing numerator and denominator of the fraction by $\cos\theta$ :

= \tan\theta \cdot \frac{\tan\theta + 1}{1 - \tan\theta}

= \frac{\tan\theta(\tan\theta + 1)}{1 - \tan\theta} = \frac{\tan^2\theta + \tan\theta}{1 - \tan\theta}

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Q4. Given that $\sin\theta = \dfrac{3}{5}$ and $\theta$ is acute, find the exact values of $\cos\theta$ and $\tan\theta$ .

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Use $\sin^2\theta + \cos^2\theta \equiv 1$ :

\cos^2\theta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}

Since $\theta$ is acute, $\cos\theta > 0$ :

\cos\theta = \frac{4}{5}

Apply $\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}$ :

\tan\theta = \frac{3/5}{4/5} = \frac{3}{4}

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Q5. Prove that $(\sin\theta + \cos\theta)^2 \equiv 1 + 2\sin\theta\cos\theta$ .

<details><summary>Show answer</summary>

Expand the LHS:

(\sin\theta + \cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta

Apply $\sin^2\theta + \cos^2\theta \equiv 1$ :

= 1 + 2\sin\theta\cos\theta = \text{RHS} \qquad \blacksquare

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Connections

Prerequisite knowledge used here:

Graphs of Trigonometric Functions — understanding where $\sin\theta$ , $\cos\theta$ , and $\tan\theta$ are positive or negative (CAST diagram) is essential when solving equations and finding all solutions in a given interval.

What these identities unlock next:

Further Trigonometric Identities — the double-angle formulae ( $\sin 2\theta$ , $\cos 2\theta$ ) are derived by applying these two identities, making this note a direct prerequisite.
Solving More Complex Trigonometric Equations — equations involving $\tan^2\theta$ are routinely simplified by substituting $\tan^2\theta \equiv \dfrac{\sin^2\theta}{\cos^2\theta}$ or using the Pythagorean identity in the form $\sec^2\theta \equiv 1 + \tan^2\theta$ (Pure Mathematics 3).
Integration and Differentiation of Trigonometric Functions — in later units, $\sin^2\theta + \cos^2\theta \equiv 1$ appears when verifying derivatives and in integration by substitution.

Introduction

Core Concept

The Quotient Identity

The Pythagorean Identity

Two Rearrangements You Must Know

Key Formulae & Definitions

Worked Examples

Example 1 — Proving an Identity

Example 2 — Solving an Equation

Example 3 — Simplifying an Expression

Common Mistakes & Examiner Pitfalls

Practice Questions

Connections

Figures

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