Trigonometric Identities notes: ICSE Class 10 Mathematics

1. Ratios recalled and what an identity is

For an acute angle $\theta$ in a right-angled triangle, $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$ , $\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$ , $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ , and $\cot\theta$ , $\sec\theta$ , $\operatorname{cosec}\theta$ are the reciprocals of $\tan\theta$ , $\cos\theta$ , $\sin\theta$ .

An identity is true for every angle for which both sides are defined, unlike an equation, which is true only for particular angles.

Example: in the $6, 8, 10$ triangle, $\sin\theta = \dfrac{3}{5}$ , $\cos\theta = \dfrac{4}{5}$ , $\tan\theta = \dfrac{3}{4}$ .

The six ratios and their values in the triangle.
Ratio	Definition	For the triangle ( $6, 8, 10$ )
$\sin\theta$	$\dfrac{\text{opposite}}{\text{hypotenuse}}$	$\dfrac{3}{5}$
$\cos\theta$	$\dfrac{\text{adjacent}}{\text{hypotenuse}}$	$\dfrac{4}{5}$
$\tan\theta$	$\dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{\sin\theta}{\cos\theta}$	$\dfrac{3}{4}$
$\cot\theta$ , $\sec\theta$ , $\operatorname{cosec}\theta$	reciprocals of $\tan\theta$ , $\cos\theta$ , $\sin\theta$	$\dfrac{4}{3}$ , $\dfrac{5}{4}$ , $\dfrac{5}{3}$

For $\theta$ at $C$: opposite $6$, adjacent $8$, hypotenuse $10$, so $\sin\theta = \dfrac{3}{5}$, $\cos\theta = \dfrac{4}{5}$.

Formulas

Tangent as a quotient
$\tan\theta = \dfrac{\sin\theta}{\cos\theta}$

Key terms

Identity: A relation true for every angle (in its range) for which both sides are defined.
Reciprocal ratios: $\operatorname{cosec}\theta = \dfrac{1}{\sin\theta}$ , $\sec\theta = \dfrac{1}{\cos\theta}$ , $\cot\theta = \dfrac{1}{\tan\theta}$ .

2. The first identity: sin² θ + cos² θ = 1

Let $AB = p$ , $BC = q$ , $AC = h$ in a triangle right-angled at $B$ . Pythagoras gives $p^2 + q^2 = h^2$ . Dividing by $h^2$ and writing $\dfrac{q}{h} = \sin A$ , $\dfrac{p}{h} = \cos A$ :

\sin^2 A + \cos^2 A = 1.

Rearranged: $\sin^2 A = 1 - \cos^2 A$ and $\cos^2 A = 1 - \sin^2 A$ .

Example: if $\sin\theta = \dfrac{3}{5}$ , then $\cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}$ and $\cos\theta = \dfrac{4}{5}$ .

Pythagoras divided by $h^2$.

All three identities hold for the $6, 8, 10$ triangle.
Identity	Left side	Right side
$\sin^2\theta + \cos^2\theta = 1$	$\dfrac{9}{25} + \dfrac{16}{25} = 1$	$1$
$1 + \tan^2\theta = \sec^2\theta$	$1 + \dfrac{9}{16} = \dfrac{25}{16}$	$\left(\dfrac{5}{4}\right)^2 = \dfrac{25}{16}$
$1 + \cot^2\theta = \operatorname{cosec}^2\theta$	$1 + \dfrac{16}{9} = \dfrac{25}{9}$	$\left(\dfrac{5}{3}\right)^2 = \dfrac{25}{9}$

Formulas

First identity
$\sin^2\theta + \cos^2\theta = 1$

Key terms

Pythagorean identity: $\sin^2\theta + \cos^2\theta = 1$ , the Pythagoras relation divided by the hypotenuse squared.

3. The other two identities

Dividing $\sin^2 A + \cos^2 A = 1$ by $\cos^2 A$ gives $1 + \tan^2 A = \sec^2 A$ ; dividing by $\sin^2 A$ gives $1 + \cot^2 A = \operatorname{cosec}^2 A$ .

Useful forms: $\tan^2 A = \sec^2 A - 1$ , $\sec^2 A - \tan^2 A = 1$ , $\cot^2 A = \operatorname{cosec}^2 A - 1$ , $\operatorname{cosec}^2 A - \cot^2 A = 1$ .

The identities hold for every acute angle and also at $0^\circ$ (for the first two); $\cot$ and $\operatorname{cosec}$ are not defined at $0^\circ$ .

Example: at $30^\circ$ : $1 + \dfrac{1}{3} = \dfrac{4}{3} = \sec^2 30^\circ$ .

The other two identities by division.

Each identity can be rearranged in several useful ways.
Identity	Useful forms
$\sin^2\theta + \cos^2\theta = 1$	$\sin^2\theta = 1 - \cos^2\theta$ , $\cos^2\theta = 1 - \sin^2\theta$
$1 + \tan^2\theta = \sec^2\theta$	$\tan^2\theta = \sec^2\theta - 1$ , $\sec^2\theta - \tan^2\theta = 1$
$1 + \cot^2\theta = \operatorname{cosec}^2\theta$	$\cot^2\theta = \operatorname{cosec}^2\theta - 1$ , $\operatorname{cosec}^2\theta - \cot^2\theta = 1$

The identities agree with the exact values of the standard angles.
$\theta$	$\sin^2\theta + \cos^2\theta$	$1 + \tan^2\theta = \sec^2\theta$	$1 + \cot^2\theta = \operatorname{cosec}^2\theta$
$30^\circ$	$\dfrac{1}{4} + \dfrac{3}{4} = 1$	$1 + \dfrac{1}{3} = \dfrac{4}{3}$	$1 + 3 = 4$
$45^\circ$	$\dfrac{1}{2} + \dfrac{1}{2} = 1$	$1 + 1 = 2$	$1 + 1 = 2$
$60^\circ$	$\dfrac{3}{4} + \dfrac{1}{4} = 1$	$1 + 3 = 4$	$1 + \dfrac{1}{3} = \dfrac{4}{3}$

Formulas

Second identity
$1 + \tan^2\theta = \sec^2\theta$
Third identity
$1 + \cot^2\theta = \operatorname{cosec}^2\theta$

Key terms

Second identity: $1 + \tan^2\theta = \sec^2\theta$ .
Third identity: $1 + \cot^2\theta = \operatorname{cosec}^2\theta$ .

4. Finding all ratios from one

Use the identity that links the given ratio to a new one, then reciprocals and $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ . Keep every value as an exact fraction and take positive roots (the angle is acute).

Example: $\sin\theta = \dfrac{28}{53}$ gives $\cos\theta = \dfrac{45}{53}$ , $\tan\theta = \dfrac{28}{45}$ , $\operatorname{cosec}\theta = \dfrac{53}{28}$ , $\sec\theta = \dfrac{53}{45}$ , $\cot\theta = \dfrac{45}{28}$ .

Range check: for an acute angle $0 < \sin\theta, \cos\theta < 1$ , $\tan\theta, \cot\theta > 0$ and $\sec\theta, \operatorname{cosec}\theta > 1$ . Values such as $\sin\theta = 1.2$ are impossible.

Pick the identity that links the given ratio to a new one.

From one ratio to all six: use $\sin^2\theta + \cos^2\theta = 1$ once.
Step	Working	Result
Find $\cos\theta$	$\cos^2\theta = 1 - \dfrac{784}{2809} = \dfrac{2025}{2809}$	$\cos\theta = \dfrac{45}{53}$
Find $\tan\theta$	$\dfrac{\sin\theta}{\cos\theta} = \dfrac{28}{45}$	$\tan\theta = \dfrac{28}{45}$
Reciprocals	turn each fraction upside down	$\operatorname{cosec}\theta = \dfrac{53}{28}$ , $\sec\theta = \dfrac{53}{45}$ , $\cot\theta = \dfrac{45}{28}$

A ratio value outside its range cannot belong to any acute angle.
For an acute angle $\theta$	Possible values	Reason
$\sin\theta$ , $\cos\theta$	between $0$ and $1$	the hypotenuse is the longest side
$\tan\theta$ , $\cot\theta$	any positive number	a leg divided by a leg
$\sec\theta$ , $\operatorname{cosec}\theta$	greater than $1$	reciprocals of numbers below $1$

Check the range before using an identity.
Claim	Test	Verdict
$\sin\theta = 1.2$	$1.2 > 1$	impossible
$\sec\theta = 0.9$	$0.9 < 1$	impossible
$\tan\theta = 7$	positive	possible
$\operatorname{cosec}\theta = 1.25$	$1.25 > 1$ , gives $\sin\theta = 0.8$	possible

Key terms

Exact fraction: A ratio kept as a fraction such as $\dfrac{45}{53}$ , never rounded.

5. Complementary angles

Angles with sum $90^\circ$ are complementary. The two acute angles of a right-angled triangle are complementary, so:

$\sin(90^\circ - \theta) = \cos\theta$ , $\cos(90^\circ - \theta) = \sin\theta$ ;
$\tan(90^\circ - \theta) = \cot\theta$ , $\cot(90^\circ - \theta) = \tan\theta$ ;
$\sec(90^\circ - \theta) = \operatorname{cosec}\theta$ , $\operatorname{cosec}(90^\circ - \theta) = \sec\theta$ .

Example: $\sin 25^\circ \div \cos 65^\circ = 1$ , because $\cos 65^\circ = \sin 25^\circ$ ; and $\tan 40^\circ\tan 50^\circ = 1$ .

Each ratio of an angle equals the co-ratio of its complement.
Complementary-angle rule	Example
$\sin(90^\circ - \theta) = \cos\theta$	$\sin 70^\circ = \cos 20^\circ$
$\cos(90^\circ - \theta) = \sin\theta$	$\cos 70^\circ = \sin 20^\circ$
$\tan(90^\circ - \theta) = \cot\theta$	$\tan 70^\circ = \cot 20^\circ$
$\cot(90^\circ - \theta) = \tan\theta$	$\cot 70^\circ = \tan 20^\circ$
$\sec(90^\circ - \theta) = \operatorname{cosec}\theta$	$\sec 70^\circ = \operatorname{cosec} 20^\circ$
$\operatorname{cosec}(90^\circ - \theta) = \sec\theta$	$\operatorname{cosec} 70^\circ = \sec 20^\circ$

The two acute angles add to $90^\circ$: $\sin A = \dfrac{BC}{AC} = \cos C$.

Each term is $1$, so the expression is $1 + 1 - 1 = 1$.
Term	Complementary rule	Value
$\dfrac{\sin 25^\circ}{\cos 65^\circ}$	$\cos 65^\circ = \sin 25^\circ$	$1$
$\tan 40^\circ\tan 50^\circ$	$\tan 50^\circ = \cot 40^\circ$ , so $\tan 40^\circ\cot 40^\circ$	$1$
$\sec 70^\circ\sin 20^\circ$	$\sin 20^\circ = \cos 70^\circ$ , so $\sec 70^\circ\cos 70^\circ$	$1$

Formulas

Complementary rule
$\sin(90^\circ - \theta) = \cos\theta$

Key terms

Complementary angles: Two angles whose sum is $90^\circ$ .
Co-ratio: The partner of a ratio: sin and cos, tan and cot, sec and cosec.

6. Proving identities

Work on the more complicated side only; never move terms across the equals sign.
Write every ratio with $\sin\theta$ and $\cos\theta$ .
Combine fractions, or multiply by a conjugate such as $1 - \cos\theta$ for $1 + \cos\theta$ .
Use $\sin^2\theta + \cos^2\theta = 1$ (and its forms) to simplify.
Check at $30^\circ$ , $45^\circ$ or $60^\circ$ to catch slips.

Example: $\sec\theta - \cos\theta = \dfrac{1 - \cos^2\theta}{\cos\theta} = \dfrac{\sin^2\theta}{\cos\theta} = \sin\theta\tan\theta$ .

Work on one side only, until it matches the other.

Convert first; the identity usually appears by itself.
Ratio	In sine and cosine
$\tan\theta$	$\dfrac{\sin\theta}{\cos\theta}$
$\cot\theta$	$\dfrac{\cos\theta}{\sin\theta}$
$\sec\theta$	$\dfrac{1}{\cos\theta}$
$\operatorname{cosec}\theta$	$\dfrac{1}{\sin\theta}$
$\tan\theta + \cot\theta$	$\dfrac{1}{\sin\theta\cos\theta}$

Proof by simplifying the left side only.
Step	Working
Use $\sec\theta = \dfrac{1}{\cos\theta}$	$\sec\theta - \cos\theta = \dfrac{1}{\cos\theta} - \cos\theta$
Common denominator	$= \dfrac{1 - \cos^2\theta}{\cos\theta}$
Identity	$= \dfrac{\sin^2\theta}{\cos\theta}$
Split	$= \sin\theta \times \dfrac{\sin\theta}{\cos\theta} = \sin\theta\tan\theta$
Check at $30^\circ$	$\dfrac{2}{\sqrt{3}} - \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{6} = \dfrac{1}{2} \times \dfrac{\sqrt{3}}{3}$

The conjugate denominator turns $1 - \sin^2\theta$ into $\cos^2\theta$.
Step	Working
Common denominator	$\dfrac{(1 - \sin\theta) + (1 + \sin\theta)}{(1 + \sin\theta)(1 - \sin\theta)}$
Simplify	$= \dfrac{2}{1 - \sin^2\theta}$
Identity	$= \dfrac{2}{\cos^2\theta} = 2\sec^2\theta$
Check at $45^\circ$	$\dfrac{1}{1 + \frac{\sqrt{2}}{2}} + \dfrac{1}{1 - \frac{\sqrt{2}}{2}} = 4 = 2 \times 2$

Multiplying by the conjugate $1 - \cos\theta$ gives $\sin^2\theta$.
Step	Working
Multiply top and bottom by $1 - \cos\theta$	$\dfrac{\sin\theta(1 - \cos\theta)}{(1 + \cos\theta)(1 - \cos\theta)}$
Denominator	$= \dfrac{\sin\theta(1 - \cos\theta)}{1 - \cos^2\theta} = \dfrac{\sin\theta(1 - \cos\theta)}{\sin^2\theta}$
Cancel $\sin\theta$	$= \dfrac{1 - \cos\theta}{\sin\theta}$
Check at $60^\circ$	$\dfrac{\frac{\sqrt{3}}{2}}{\frac{3}{2}} = \dfrac{\sqrt{3}}{3} = \dfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

Formulas

Conjugate pair
$(1 - \sin\theta)(1 + \sin\theta) = \cos^2\theta$

Key terms

Conjugate: The expression with the opposite sign between the same two terms, e.g. $1 + \sin\theta$ and $1 - \sin\theta$ .

7. Simplifying and evaluating expressions

Simplifying: spot an identity inside the expression. $(1 + \tan^2\theta)(1 - \sin\theta)(1 + \sin\theta) = \sec^2\theta\cos^2\theta = 1$ .
Evaluating with a given ratio: find the other ratios as exact fractions, then substitute. If $\sin\theta = \dfrac{12}{37}$ , then $\sec\theta + \tan\theta = \dfrac{37}{35} + \dfrac{12}{35} = \dfrac{7}{5}$ .
Standard angles: use the exact values of $30^\circ$ , $45^\circ$ , $60^\circ$ ; for other angles use complementary pairs.
Eliminating $\theta$ : from $x = 4\cos\theta$ , $y = 3\sin\theta$ , get $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ .

Four steps for evaluating with a given ratio.

Exact fractions all the way: no decimals.
Step	Working
Find $\cos\theta$	$\cos^2\theta = 1 - \dfrac{144}{1369} = \dfrac{1225}{1369}$ , so $\cos\theta = \dfrac{35}{37}$
Find $\sec\theta$ and $\tan\theta$	$\sec\theta = \dfrac{37}{35}$ , $\tan\theta = \dfrac{12}{37} \div \dfrac{35}{37} = \dfrac{12}{35}$
Add	$\sec\theta + \tan\theta = \dfrac{37 + 12}{35} = \dfrac{49}{35} = \dfrac{7}{5}$

A product of three factors collapses to the number $1$.
Step	Working
Use two identities	$(1 + \tan^2\theta) = \sec^2\theta$ and $(1 - \sin\theta)(1 + \sin\theta) = 1 - \sin^2\theta = \cos^2\theta$
Multiply	$\sec^2\theta \times \cos^2\theta = \dfrac{1}{\cos^2\theta} \times \cos^2\theta$
Result	$= 1$
Check at $60^\circ$	$(1 + 3) \times \dfrac{1}{4} = 1$

Four slips to avoid.
Common slip	Why it is wrong	Correct step
$\sin\theta + \cos\theta = 1$	only the squares add to $1$ ; at $45^\circ$ the sum is $\sqrt{2}$	$\sin^2\theta + \cos^2\theta = 1$
$\sqrt{1 - \cos^2\theta} = \sin^2\theta$	the square root gives a power one, not two	$\sqrt{1 - \cos^2\theta} = \sin\theta$
$\sin^2\theta + \sin^2(90^\circ - \theta) = 0$	complement turns sine into cosine	$= \sin^2\theta + \cos^2\theta = 1$
Moving a term across the equals sign in a proof	a proof must not assume the result	simplify one side until it equals the other

Key terms

Eliminating θ: Combining two relations in $\theta$ with an identity so that $\theta$ no longer appears.

Exam tips

How marks are awarded

Finding ratios from one (2–3 marks): identity used 1; second ratio 1; remaining ratios or answer 1.
Proof (3–4 marks): start from one side 1; algebra steps 1–2; conclusion 1.
Evaluation with complements (3 marks): each complementary conversion 1.
10-mark structured question: parts (a), (b), (c) usually 3 + 3 + 4; show all working.

Common mistakes

Writing $\sin\theta + \cos\theta = 1$ (only the squares add to $1$ ).
Taking the negative root for an acute angle.
Assuming the result and working on both sides in a proof.
Mixing a ratio with its co-ratio in complementary angles.
Rounding to decimals instead of keeping exact fractions.
Not simplifying a conjugate denominator fully.

1. Ratios recalled and what an identity is

Formulas

Key terms

2. The first identity: sin² θ + cos² θ = 1

Formulas

Key terms

3. The other two identities

Formulas

Key terms

4. Finding all ratios from one

Key terms

5. Complementary angles

Formulas

Key terms

6. Proving identities

Formulas

Key terms

7. Simplifying and evaluating expressions

Key terms

Exam tips

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