You already define all six trigonometric ratios from the coordinates x, y and r. This unit links the ratios to one another. An identity is an equation that is true for every angle at which both sides are defined — a small toolkit of identities lets you simplify expressions and prove new results. **Quotient identities** **Identity** | **Condition** tan A= sin Acos A | cos A!= 0 cot A= cos Asin A= 1tan A | sin A!= 0 **Pythagorean identities** **Identity** | **How it is obtained** sin2 A+cos2 A=1 | From x2+y2=r2, dividing every term by r2. 1+tan2 A=sec2 A | Divide sin2 A+cos2 A=1 by cos2 A. 1+cot2 A=cosec2 A | Divide sin2 A+cos2 A=1 by sin2 A. Every identity stays true when A is replaced by any other angle — 3theta, theta2, A+B — provided the same angle is used throughout. **Using the same angle throughout** **Correct use** | **Not an identity** sin2 3theta+cos2 3theta=1 | sintheta+costheta=1 sec2 theta2=1+tan2 theta2 | sin4theta+cos4theta=1 - Start from the more complicated side — usually the one with fractions or several terms. - Express tan, cot, sec and cosec in terms of sin and cos. - Combine fractions over a common denominator. - Factorise, then simplify with a Pythagorean identity wherever a square appears. - Stop when the side you are working on matches the other side exactly. **Common moves in a proof** **Move** | **Working** Express in sin and cos | cot Asin A = cos Asin A x sin A =cos A Pythagorean swap after factorising | (1-cos A)(1+cos A) =1-cos2 A =sin2 A Combine into one fraction | 1cos2 A-1 = 1-cos2 Acos2 A = sin2 Acos2 A =tan2 A Work down one side only. Never move terms across the equals sign or operate on both sides — the identity is exactly what you are trying to prove, so you may not assume it.

Basic Identities and Proving an Identity

Secondary 4

TGM Original Questions

From ratios to identities

You already define all six trigonometric ratios from the coordinates \(x\), \(y\) and \(r\). This unit links the ratios to one another. An identity is an equation that is true for every angle at which both sides are defined — a small toolkit of identities lets you simplify expressions and prove new results.

Quotient identities

Identity	Condition
\(\tan A=\dfrac{\sin A}{\cos A}\)	\(\cos A\ne 0\)
\(\cot A=\dfrac{\cos A}{\sin A}=\dfrac{1}{\tan A}\)	\(\sin A\ne 0\)

Pythagorean identities

Identity	How it is obtained
\(\sin^2 A+\cos^2 A=1\)	From \(x^2+y^2=r^2\), dividing every term by \(r^2\).
\(1+\tan^2 A=\sec^2 A\)	Divide \(\sin^2 A+\cos^2 A=1\) by \(\cos^2 A\).
\(1+\cot^2 A=\operatorname{cosec}^2 A\)	Divide \(\sin^2 A+\cos^2 A=1\) by \(\sin^2 A\).

Every identity stays true when \(A\) is replaced by any other angle — \(3\theta\), \(\dfrac{\theta}{2}\), \(A+B\) — provided the same angle is used throughout.

Using the same angle throughout

Correct use	Not an identity
\(\sin^2 3\theta+\cos^2 3\theta=1\)	\(\sin\theta+\cos\theta=1\)
\(\sec^2\dfrac{\theta}{2}=1+\tan^2\dfrac{\theta}{2}\)	\(\sin^4\theta+\cos^4\theta=1\)

Method: Proving an identity

Start from the more complicated side — usually the one with fractions or several terms.
Express \(\tan\), \(\cot\), \(\sec\) and \(\operatorname{cosec}\) in terms of \(\sin\) and \(\cos\).
Combine fractions over a common denominator.
Factorise, then simplify with a Pythagorean identity wherever a square appears.
Stop when the side you are working on matches the other side exactly.

Common moves in a proof

Move	Working
Express in \(\sin\) and \(\cos\)	\(\cot A\sin A\) \(=\dfrac{\cos A}{\sin A}\times\sin A\) \(=\cos A\)
Pythagorean swap after factorising	\((1-\cos A)(1+\cos A)\) \(=1-\cos^2 A\) \(=\sin^2 A\)
Combine into one fraction	\(\dfrac{1}{\cos^2 A}-1\) \(=\dfrac{1-\cos^2 A}{\cos^2 A}\) \(=\dfrac{\sin^2 A}{\cos^2 A}\) \(=\tan^2 A\)

Exam check

Work down one side only. Never move terms across the equals sign or operate on both sides — the identity is exactly what you are trying to prove, so you may not assume it.

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