MathsICSE Class 10 · Trigonometric Identities

The Three Fundamental Trigonometric Identities: Derivation and Uses (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

The three fundamental identities are $\sin^2 A + \cos^2 A = 1$, $1 + \tan^2 A = \sec^2 A$ and $1 + \cot^2 A = \operatorname{cosec}^2 A$. All three come from Pythagoras' theorem in a right-angled triangle, hold for every angle where the ratios are defined, and let you find all ratios from one.

Divide Pythagoras' equation by each squared side in turn, and each division gives one of the three identities.

Illustration drawn for QuestionBuddy.

Understand The Three Fundamental Trigonometric Identities

These are exactly the identities named in the CISCE syllabus for Class 10 ( $0^{\circ} \leq A \leq 9 0^{\circ}$ ).

Derivation. In $△ A B C$ , right-angled at $B$ , let the sides be $A B$ (adjacent to $A$ ), $B C$ (opposite) and $A C$ (hypotenuse). Pythagoras gives $A B^{2} + B C^{2} = A C^{2}$ .

1. Divide by $A C^{2}$ : $(\frac{A B}{A C})^{2} + (\frac{B C}{A C})^{2} = 1$ , so $cos^{2} A + sin^{2} A = 1$ .
2. Divide by $A B^{2}$ : $1 + (\frac{B C}{A B})^{2} = (\frac{A C}{A B})^{2}$ , so $1 + tan^{2} A = sec^{2} A$ ( $A \neq = 9 0^{\circ}$ ).
3. Divide by $B C^{2}$ : $(\frac{A B}{B C})^{2} + 1 = (\frac{A C}{B C})^{2}$ , so $1 + cot^{2} A = cosec^{2} A$ ( $A \neq = 0^{\circ}$ ).

Rearranged forms (learn these as well):

$sin^{2} A = 1 - cos^{2} A$ , $cos^{2} A = 1 - sin^{2} A$ .
$sec^{2} A - tan^{2} A = 1$ , $tan^{2} A = sec^{2} A - 1$ .
$cosec^{2} A - cot^{2} A = 1$ , $cot^{2} A = cosec^{2} A - 1$ .

Note the minus sign: $cosec^{2} A + cot^{2} A = 1$ is false (2024 Q1(xi)).

Uses.

Finding all ratios from one. For acute $A$ , every ratio is positive, so take the positive square root. Example: $sin A = \frac{3}{5}$ gives $cos A = 1 - \frac{9}{25} = \frac{4}{5}$ .
Simplifying expressions such as $(1 + cot^{2} A) sin^{2} A = 1$ .
Using a given relation: if a condition looks like part of an identity (for example $cos A = 1 - cos^{2} A$ ), replace it by the other ratio ( $sin^{2} A$ ). This was the idea behind 2025 Q1(x).
Eliminating the angle: if $x = a sec A$ and $y = b tan A$ , then $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = sec^{2} A - tan^{2} A = 1$ .

Range. For $0^{\circ} \leq A \leq 9 0^{\circ}$ : $0 \leq sin A, cos A \leq 1$ ; $sec A \geq 1$ and $cosec A \geq 1$ where defined.

The Three Fundamental Trigonometric Identities: formulas to remember

Identity 1: $sin^{2} A + cos^{2} A = 1$; Divide Pythagoras by the hypotenuse squared.
Identity 2: $1 + tan^{2} A = sec^{2} A$; Divide by the adjacent side squared; $A \neq = 9 0^{\circ}$ .
Identity 3: $1 + cot^{2} A = cosec^{2} A$; Divide by the opposite side squared; $A \neq = 0^{\circ}$ .
Difference forms: $sec^{2} A - tan^{2} A = 1, cosec^{2} A - cot^{2} A = 1$; Minus, not plus.

How to solve The Three Fundamental Trigonometric Identities questions

Match what you are given to one identity, rearrange it, and take the positive root for an acute angle.

1Identify the pair of ratios linked by one identity: $sin$ – $cos$ , $tan$ – $sec$ or $cot$ – $cosec$ .
2Rearrange to make the unknown ratio the subject and substitute.
3Take the positive square root (acute angle) and simplify.
4Get the remaining ratios from quotient and reciprocal relations.
5For elimination, make $sec A$ , $tan A$ (or $sin A$ , $cos A$ ) the subjects, square and use the identity.

The Three Fundamental Trigonometric Identities in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

sin² + cos² = 1 with a kite string

Suppose a kite string 50 m long is pulled straight, and the angle $A$ it makes with the ground has $cos A = 0.28$ .

How high is the kite, and how far away horizontally?

Try it first, then show the solutionSolution

Horizontal distance $= 50 cos A = 14$ m.
$sin A = 1 - 0.2 8^{2} = 0.9216 = 0.96$ , so height $= 50 \times 0.96 = 48$ m.

Answer: 48 m high, 14 m away

Memory hook: Know cos, get sin: $1 - cos^{2}$ .

1 + tan² = sec² for a ramp

Imagine a wheelchair ramp 5 m long whose angle $A$ with the ground has $sec A = \frac{25}{24}$ .

Find $tan A$ , the height the ramp rises and the horizontal distance it covers.

Try it first, then show the solutionSolution

$tan A = \frac{625}{576} - 1 = \frac{7}{24}$ ; $cos A = \frac{24}{25}$ , $sin A = \frac{7}{25}$ .
Height $= 5 \times \frac{7}{25} = 1.4$ m; horizontal $= 5 \times \frac{24}{25} = 4.8$ m.

Answer: $tan A = \frac{7}{24}$ ; rises 1.4 m over 4.8 m

Memory hook: sec and tan travel together.

Tricks to remember The Three Fundamental Trigonometric Identities

Short memory aids. Say each one aloud once, then use it on a question the same day.

Pythagoras in three disguises
Divide $a^{2} + b^{2} = c^{2}$ by each side squared in turn: hypotenuse → sin/cos, adjacent → tan/sec, opposite → cot/cosec.
Use it when: Rebuilding the identities if you forget them.
Partners
sin with cos, tan with sec, cot with cosec: the 'co' ratios pair with the 'co' ratios.
Use it when: Picking the right identity.
Sec and cosec are the big ones
$sec^{2}$ and $cosec^{2}$ are always 1 more than their partners, so they sit alone on one side.
Use it when: Avoiding the plus-sign error.

The Three Fundamental Trigonometric Identities: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: One identity, rearranged · 1 mark · MCQBoard level

$\frac{1}{1 + t a n ^{2} A}$ equals:
(a) $sin^{2} A$
(b) $cos^{2} A$
(c) $sec^{2} A$
(d) $tan^{2} A$

Try it first, then show the solutionSolution

$1 + tan^{2} A = sec^{2} A$ , so $\frac{1}{s e c ^{2} A} = cos^{2} A$ .

Answer: (b)

Example 2: All ratios from sin A · 3 marks · Short answerBoard level

If $sin A = \frac{8}{17}$ and $A$ is acute, use identities to find $cos A$ , $tan A$ and $sec A$ .

Try it first, then show the solutionSolution

$cos A = 1 - \frac{64}{289} = \frac{225}{289} = \frac{15}{17}$ .
$tan A = \frac{8/17}{15/17} = \frac{8}{15}$ ; $sec A = \frac{17}{15}$ .
Check: $1 + \frac{64}{225} = \frac{289}{225} = sec^{2} A$ ✓

Answer: $cos A = \frac{15}{17}$ , $tan A = \frac{8}{15}$ , $sec A = \frac{17}{15}$

Example 3: From cot A to an expression · 3 marks · Short answerHarder board level

If $cot A = \frac{20}{21}$ and $A$ is acute, find $cosec A$ , $sin A$ , $cos A$ , and the value of $\frac{s i n A + c o s A}{s i n A - c o s A}$ .

Try it first, then show the solutionSolution

$cosec^{2} A = 1 + \frac{400}{441} = \frac{841}{441}$ , so $cosec A = \frac{29}{21}$ and $sin A = \frac{21}{29}$ .
$cos A = cot A \times sin A = \frac{20}{21} \times \frac{21}{29} = \frac{20}{29}$ .
$\frac{s i n A + c o s A}{s i n A - c o s A} = \frac{41/29}{1/29} = 41$ .

Answer: $cosec A = \frac{29}{21}$ , $sin A = \frac{21}{29}$ , $cos A = \frac{20}{29}$ ; value 41

Example 4: Eliminating the angle · 3 marks · Short answerHarder board level

If $x = 3 sec A$ and $y = 4 tan A$ , find a relation between $x$ and $y$ that does not involve $A$ . Check it when $A = 6 0^{\circ}$ .

Try it first, then show the solutionSolution

$sec A = \frac{x}{3}$ and $tan A = \frac{y}{4}$ .
$sec^{2} A - tan^{2} A = 1$ gives $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ .
Check at $6 0^{\circ}$ : $x = 6$ , $y = 43$ ; $\frac{36}{9} - \frac{48}{16} = 4 - 3 = 1$ ✓

Answer: $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$

Example 5: Using a given condition · 3 marks · Short answerHOTS / competency

If $cos A + cos^{2} A = 1$ , prove that $sin^{2} A + sin^{4} A = 1$ .

Try it first, then show the solutionSolution

From the condition, $cos A = 1 - cos^{2} A = sin^{2} A$ .
So $sin^{4} A = (sin^{2} A)^{2} = cos^{2} A$ .
$sin^{2} A + sin^{4} A = sin^{2} A + cos^{2} A = 1$ .

Answer: Proved

Example 6: Case study: a playground slide · 4 marks · Case-basedHOTS / competency

A straight slide 6.5 m long makes an angle $A$ with the ground, where $sec A = \frac{13}{12}$ .
(a) Use an identity to find $tan A$ , then find $cos A$ and $sin A$ .
(b) Find the height of the top of the slide and the horizontal distance it covers.
(c) Show that your answers in (b) satisfy Pythagoras' theorem, and explain the link with $sin^{2} A + cos^{2} A = 1$ .

Try it first, then show the solutionSolution

(a) $tan^{2} A = \frac{169}{144} - 1 = \frac{25}{144}$ , so $tan A = \frac{5}{12}$ ; $cos A = \frac{12}{13}$ , $sin A = tan A cos A = \frac{5}{13}$ .
(b) Height $= 6.5 \times \frac{5}{13} = 2.5$ m; horizontal $= 6.5 \times \frac{12}{13} = 6$ m.
(c) $2. 5^{2} + 6^{2} = 6.25 + 36 = 42.25 = 6. 5^{2}$ ✓. Dividing by $6. 5^{2}$ gives $sin^{2} A + cos^{2} A = 1$ : the identity is Pythagoras scaled to a hypotenuse of 1.

Answer: (a) $\frac{5}{12}$ , $\frac{12}{13}$ , $\frac{5}{13}$ (b) 2.5 m and 6 m

Where students lose marks in The Three Fundamental Trigonometric Identities

The slips examiners see most often, and what to write instead.

Mistake: Wrong sign in identity 3
$cosec^{2} A$ is 1 more than $cot^{2} A$ ; their sum is not 1.
Wrong working
$cosec^{2} θ + cot^{2} θ = 1$
Right working
$cosec^{2} θ - cot^{2} θ = 1$
Do this: Derive it from $1 + cot^{2} A = cosec^{2} A$ when in doubt.
Usually costs: Reported in 2024 Q1(xi): candidates accepted the false statement.
Mistake: Not linking a condition to an identity
A given relation often contains half of an identity.
Wrong working
Trying to solve $sin^{2} A + sin A = 1$ for $A$ first.
Right working
$sin A = 1 - sin^{2} A = cos^{2} A$ , then substitute.
Do this: Ask: which identity does the given expression resemble?
Usually costs: Reported in 2025 Q1(x): candidates could not relate the assertion to the reason.
Mistake: Negative or missing square roots
For an acute angle every ratio is positive.
Wrong working
$cos A = \pm \frac{15}{17}$ left unresolved.
Right working
$A$ acute, so $cos A = \frac{15}{17}$ .
Do this: State ' $A$ acute, so positive root'.
Usually costs: The accuracy mark.

Exam tips

2024 Q1(xi) (1 mark): only $sin^{2} θ + cos^{2} θ = 1$ was valid; $cosec^{2} θ + cot^{2} θ = 1$ was the trap.
2025 Q1(x) (1 mark): from $sin^{2} A + sin A = 1$ , $sin A = cos^{2} A$ , which gives $cos^{4} A + cos^{2} A = 1$ ; (R) $1 - sin^{2} A = cos^{2} A$ is the correct reason.
When asked to find ratios 'using identities', show the identity line; drawing a triangle alone may not earn the method mark.
Write answers as fractions in lowest terms unless decimals are given.

Practice questions on The Three Fundamental Trigonometric Identities

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
$(1 + cot^{2} A) sin^{2} A$ equals:
(a) 0
(b) 1
(c) $cos^{2} A$
(d) $tan^{2} A$
Hint
$1 + cot^{2} A = cosec^{2} A$ .
Answer
(b)
Q2 · 2 marks · Short answerBoard level
If $cos A = 0.28$ and $A$ is acute, find $sin A$ , $tan A$ and $cot A$ .
Hint
$sin A = 1 - 0.0784$ .
Answer
$sin A = 0.96$ , $tan A = \frac{24}{7}$ , $cot A = \frac{7}{24}$ .
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): There is an acute angle $A$ with $sin A = 0.6$ and $cos A = 0.6$ .
Reason (R): $sin^{2} A + cos^{2} A = 1$ for every angle $A$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
$0.36 + 0.36 = ?$
Answer
(b)
Q4 · 3 marks · Short answerHarder board level
If $sec A = x + \frac{1}{4 x}$ , prove that $sec A + tan A = 2 x$ or $\frac{1}{2 x}$ .
Hint
$tan^{2} A = sec^{2} A - 1$ is a perfect square.
Answer
$tan^{2} A = x^{2} + \frac{1}{2} + \frac{1}{16 x ^{2}} - 1 = (x - \frac{1}{4 x})^{2}$ , so $tan A = \pm (x - \frac{1}{4 x})$ . Then $sec A + tan A = 2 x$ or $\frac{1}{2 x}$ .
Q5 · 3 marks · Short answerHarder board level
If $tan A = \frac{1}{7}$ , find the value of $\frac{cosec ^{2} A - s e c ^{2} A}{cosec ^{2} A + s e c ^{2} A}$ .
Hint
$cosec^{2} A = 1 + cot^{2} A = 8$ , $sec^{2} A = 1 + tan^{2} A = \frac{8}{7}$ .
Answer
$\frac{8 - \frac{8}{7}}{8 + \frac{8}{7}} = \frac{48}{64} = \frac{3}{4}$
Q6 · 2 marks · Short answerHarder board level
If $x = 5 cos A$ and $y = 3 sin A$ , eliminate $A$ .
Hint
Make $cos A$ and $sin A$ the subjects.
Answer
$\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = cos^{2} A + sin^{2} A = 1$
Q7 · 3 marks · Short answerHOTS / competency
If $3 sin^{2} A + 7 cos^{2} A = 4$ and $A$ is acute, find $tan A$ and hence $A$ .
Hint
Write $3 sin^{2} A = 3 - 3 cos^{2} A$ .
Answer
$3 + 4 cos^{2} A = 4$ , so $cos^{2} A = \frac{1}{4}$ , $cos A = \frac{1}{2}$ , $sin A = \frac{3}{2}$ ; $tan A = 3$ and $A = 6 0^{\circ}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the three identities.Show answer
$sin^{2} A + cos^{2} A = 1$ , $1 + tan^{2} A = sec^{2} A$ , $1 + cot^{2} A = cosec^{2} A$
2. $sin A = \frac{3}{5}$ , $A$ acute. $cos A$ ?Show answer
$\frac{4}{5}$
3. $sec^{2} A - tan^{2} A$ ?Show answer
1
4.Which side do you divide Pythagoras by to get identity 2?Show answer
The side adjacent to $A$ (squared).
5.Is $cosec^{2} A + cot^{2} A = 1$ true?Show answer
No; $cosec^{2} A - cot^{2} A = 1$ .

The Three Fundamental Trigonometric Identities: frequently asked questions

What are the three fundamental trigonometric identities?

sin²A + cos²A = 1, 1 + tan²A = sec²A and 1 + cot²A = cosec²A. They hold for all angles where the ratios are defined; ICSE uses them for 0° ≤ A ≤ 90°.

How are the identities derived?

Start from Pythagoras' theorem in a right-angled triangle and divide by the square of the hypotenuse, of the adjacent side, or of the opposite side.

How do I find cos A if sin A is given?

Use cos A = √(1 − sin²A), taking the positive root when A is acute. Then tan A = sin A ÷ cos A, and the reciprocals give sec, cosec and cot.