MathsICSE Class 10 · Trigonometric Identities

How to Prove Trigonometric Identities: Methods and Worked Proofs (ICSE Class 10)

Q: What is the best way to start proving a trigonometric identity?

Start with the more complicated side. If it has many different ratios, convert everything to sine and cosine; if it has fractions, combine them; if a denominator is 1 ± sin A or sec A ± tan A, multiply by the conjugate.

Q: Can I work on both sides of an identity?

Yes, if you simplify each side separately to the same expression and then write LHS = RHS. Do not move terms from one side to the other as in an equation.

Q: Which identities are in the ICSE Class 10 syllabus?

sin²A + cos²A = 1, 1 + tan²A = sec²A and 1 + cot²A = cosec²A, for 0° ≤ A ≤ 90°, used to prove and simplify expressions.

Updated 15 October 2026 · Fact-checked

To prove a trigonometric identity, start from the more complicated side and transform it, step by step, into the other side using $\sin^2 A + \cos^2 A = 1$, $1 + \tan^2 A = \sec^2 A$ and $1 + \cot^2 A = \operatorname{cosec}^2 A$. Converting to sine and cosine, combining fractions and factorising usually finish the job.

Start from one side, convert to sin and cos, combine and factorise using the three identities, until it matches the other side.

Illustration drawn for QuestionBuddy.

Understand Proving Trigonometric Identities

The CISCE syllabus asks you to 'use identities to solve/prove simple algebraic trigonometric expressions' for $0^{\circ} \leq A \leq 9 0^{\circ}$ , using the three identities. A 4-mark proof appeared in both the 2024 and 2025 papers (2024 Q3(i), 2025 Q4(iii)).

The rules of a proof.

Work on one side (usually the longer or more complicated one) until it becomes the other side. Do not move terms across the '=' sign as in an equation.
Alternatively, simplify LHS and RHS separately to the same expression, then write 'LHS = RHS'.
Every line must follow from the one before; write the identity you use.

Toolkit.

1. **Convert to $sin$ and $cos$ .** $tan A = \frac{s i n A}{c o s A}$ , $sec A = \frac{1}{c o s A}$ and so on; then combine fractions.
2. Combine fractions. Take the LCM, and look for $sin^{2} A + cos^{2} A$ in the numerator.
3. Conjugates. Multiply by $1 - sin A$ , $sec A + tan A$ , etc., to use $(a - b) (a + b) = a^{2} - b^{2}$ .
4. Pairs that multiply to 1. $(sec A - tan A) (sec A + tan A) = 1$ and $(cosec A - cot A) (cosec A + cot A) = 1$ , so $\frac{1}{s e c A - t a n A} = sec A + tan A$ .
5. Replace 1 cleverly. Write $1 = sec^{2} A - tan^{2} A$ or $1 = sin^{2} A + cos^{2} A$ to create a common factor.
6. Algebraic identities. $a^{3} + b^{3}$ , $a^{2} - b^{2}$ and $(a + b)^{2}$ often appear; see the separate topic on algebraic identities.

Choosing a starting move.

Many different ratios → convert to $sin$ and $cos$ .
Two fractions added → common denominator.
$1 \pm sin A$ or $sec A \pm tan A$ in a denominator → multiply by the conjugate.
A square root → rewrite what is inside as a perfect square.

Checking. Before writing a proof, test it with $A = 3 0^{\circ}$ or $6 0^{\circ}$ ; if the two sides differ, you have misread it. The check is not itself a proof.

Domain. Avoid angles that make a denominator zero (for example $tan A = 1$ in $\frac{1}{1 - t a n A}$ ).

Proving Trigonometric Identities: formulas to remember

The three identities: $sin^{2} A + cos^{2} A = 1, 1 + tan^{2} A = sec^{2} A, 1 + cot^{2} A = cosec^{2} A$; $0^{\circ} \leq A \leq 9 0^{\circ}$ (where defined).
Product pairs: $(sec A - tan A) (sec A + tan A) = 1$; Likewise for $cosec A \pm cot A$ .
Quotient and reciprocal relations: $tan A = \frac{s i n A}{c o s A}, cot A = \frac{c o s A}{s i n A}, sec A = \frac{1}{c o s A}, cosec A = \frac{1}{s i n A}$; Convert before combining.

How to solve Proving Trigonometric Identities questions

Pick the harder side, choose one move from the toolkit, and keep going until it matches the other side.

1Check the identity numerically at $3 0^{\circ}$ or $6 0^{\circ}$ (in rough work).
2Choose the more complicated side and write 'LHS =' (or 'RHS =').
3Make one move at a time: convert to $sin$ / $cos$ , combine fractions, multiply by a conjugate, or factorise.
4Use an identity whenever $sin^{2} + cos^{2}$ , $sec^{2} - tan^{2}$ or $cosec^{2} - cot^{2}$ appears.
5Stop when you reach the other side exactly, and write '= RHS. Hence proved.'

Proving 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.

Finding other ratios from one

Suppose a ramp makes an angle $A$ with the ground and a phone app shows only $sin A = 0.28$ .

Find $cos A$ and $tan A$ without finding $A$ .

Try it first, then show the solutionSolution

$cos^{2} A = 1 - 0.2 8^{2} = 1 - 0.0784 = 0.9216$ , so $cos A = 0.96$ ( $A$ acute).
$tan A = \frac{0.28}{0.96} = \frac{7}{24}$ .

Answer: $cos A = 0.96$ , $tan A = \frac{7}{24} \approx 0.29$

Memory hook: One ratio + identity = all ratios.

Checking an identity numerically

Imagine you have proved that $\frac{1}{s e c A - t a n A} - \frac{1}{c o s A} = tan A$ and want a quick check with a calculator at $A = 4 0^{\circ}$ .

Do the two sides agree?

Try it first, then show the solutionSolution

$sec 4 0^{\circ} \approx 1.3054$ and $tan 4 0^{\circ} \approx 0.8391$ , so $\frac{1}{1.3054 - 0.8391} - 1.3054 \approx 2.1445 - 1.3054 = 0.8391$ .
This equals $tan 4 0^{\circ} \approx 0.8391$ .

Answer: Yes, both sides are about 0.8391.

Memory hook: Test with a number; prove with algebra.

Tricks to remember Proving Trigonometric Identities

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

SC first
Stuck? Turn everything into Sine and Cosine.
Use it when: Mixed-ratio identities.
1 is a disguise
1 can be $sin^{2} + cos^{2}$ , $sec^{2} - tan^{2}$ or $cosec^{2} - cot^{2}$ ; choose the one that factorises.
Use it when: Proofs like $\frac{s i n A - c o s A + 1}{s i n A + c o s A - 1}$ .
Flip the pair
$\frac{1}{s e c A - t a n A} = sec A + tan A$ (and the same for cosec, cot).
Use it when: Shortening many proofs to one line.

Proving Trigonometric Identities: solved examples

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

Example 1: Adding conjugate fractions · 1 mark · MCQBoard level

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

Try it first, then show the solutionSolution

$\frac{( 1 - s i n A ) + ( 1 + s i n A )}{1 - s i n ^{2} A} = \frac{2}{c o s ^{2} A} = 2 sec^{2} A$ .

Answer: (a)

Example 2: Cross-multiplication form · 2 marks · Short answerBoard level

Prove that $\frac{c o s A}{1 - s i n A} = \frac{1 + s i n A}{c o s A}$ .

Try it first, then show the solutionSolution

LHS $= \frac{c o s A}{1 - s i n A} \times \frac{1 + s i n A}{1 + s i n A} = \frac{c o s A ( 1 + s i n A )}{1 - s i n ^{2} A}$ .
$= \frac{c o s A ( 1 + s i n A )}{c o s ^{2} A} = \frac{1 + s i n A}{c o s A} =$ RHS.

Answer: Proved

Example 3: Using the product pair · 3 marks · Short answerHarder board level

Prove that $\frac{1}{s e c A - t a n A} - \frac{1}{c o s A} = \frac{1}{c o s A} - \frac{1}{s e c A + t a n A}$ .

Try it first, then show the solutionSolution

$(sec A - tan A) (sec A + tan A) = sec^{2} A - tan^{2} A = 1$ , so $\frac{1}{s e c A - t a n A} = sec A + tan A$ and $\frac{1}{s e c A + t a n A} = sec A - tan A$ .
LHS $= sec A + tan A - sec A = tan A$ .
RHS $= sec A - (sec A - tan A) = tan A$ . So LHS = RHS.

Answer: Proved (both sides equal $tan A$ )

Example 4: Replacing 1 by sec² − tan² · 4 marks · Long answerHarder board level

Prove that $\frac{s i n A - c o s A + 1}{s i n A + c o s A - 1} = \frac{1}{s e c A - t a n A}$ .

Try it first, then show the solutionSolution

Divide numerator and denominator by $cos A$ : LHS $= \frac{t a n A - 1 + s e c A}{t a n A + 1 - s e c A}$ .
In the numerator write $1 = sec^{2} A - tan^{2} A$ : $(tan A + sec A) - (sec A - tan A) (sec A + tan A) = (sec A + tan A) (1 - sec A + tan A)$ .
The bracket $(1 - sec A + tan A)$ is the denominator, so LHS $= sec A + tan A$ .
$sec A + tan A = \frac{1}{s e c A - t a n A}$ (product pair) $=$ RHS.

Answer: Proved

Example 5: A factor that cancels · 4 marks · Long answerHOTS / competency

Prove that $\frac{c o s A}{1 - t a n A} + \frac{s i n A}{1 - c o t A} = sin A + cos A$ (for $A \neq = 4 5^{\circ}$ ).

Try it first, then show the solutionSolution

$\frac{c o s A}{1 - \frac{s i n A}{c o s A}} = \frac{c o s ^{2} A}{c o s A - s i n A}$ and $\frac{s i n A}{1 - \frac{c o s A}{s i n A}} = \frac{s i n ^{2} A}{s i n A - c o s A}$ .
LHS $= \frac{c o s ^{2} A}{c o s A - s i n A} - \frac{s i n ^{2} A}{c o s A - s i n A} = \frac{c o s ^{2} A - s i n ^{2} A}{c o s A - s i n A}$ .
$= \frac{( c o s A - s i n A ) ( c o s A + s i n A )}{c o s A - s i n A} = sin A + cos A =$ RHS.

Answer: Proved

Example 6: Case study: two methods, one identity · 4 marks · Case-basedHOTS / competency

Two students are asked to prove $(sec A - cos A) (cosec A - sin A) = \frac{1}{t a n A + c o t A}$ .
(a) Riya simplifies the LHS. Show her working.
(b) Arjun simplifies the RHS. Show his working.
(c) Check the identity at $A = 6 0^{\circ}$ using exact values.

Try it first, then show the solutionSolution

(a) $sec A - cos A = \frac{1 - c o s ^{2} A}{c o s A} = \frac{s i n ^{2} A}{c o s A}$ ; $cosec A - sin A = \frac{c o s ^{2} A}{s i n A}$ . Product $= sin A cos A$ .
(b) $tan A + cot A = \frac{s i n ^{2} A + c o s ^{2} A}{s i n A c o s A} = \frac{1}{s i n A c o s A}$ , so RHS $= sin A cos A$ . Both sides equal $sin A cos A$ .
(c) LHS $= (2 - \frac{1}{2}) (\frac{2}{3} - \frac{3}{2}) = \frac{3}{2} \times \frac{1}{2 3} = \frac{3}{4}$ ; RHS $= sin 6 0^{\circ} cos 6 0^{\circ} = \frac{3}{4}$ ✓

Answer: Both sides reduce to $sin A cos A$ ; at $6 0^{\circ}$ each is $\frac{3}{4}$ .

Where students lose marks in Proving Trigonometric Identities

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

Mistake: Treating the identity like an equation
Moving terms across or squaring both sides assumes what you are proving.
Wrong working
Cross-multiplying LHS and RHS and simplifying both together.
Right working
Transform one side until it becomes the other.
Do this: Start with 'LHS =' and never write the RHS until the last line.
Usually costs: Method marks in a 3- or 4-mark proof.
Mistake: Slips with the common denominator
Many proofs need the LCM $sin A cos A$ ; signs and squares go wrong easily.
Wrong working
$cot A + tan A - 1 = \frac{c o s A + s i n A - 1}{s i n A c o s A}$
Right working
$cot A + tan A - 1 = \frac{c o s ^{2} A + s i n ^{2} A - s i n A c o s A}{s i n A c o s A}$
Do this: Write each term over the LCM separately before adding.
Usually costs: Reported in 2025 Q4(iii) and 2024 Q3(i): errors after taking the LCM.
Mistake: Not using an algebraic identity
Sums of cubes and differences of squares unlock many proofs.
Wrong working
Expanding $sin^{3} A + cos^{3} A$ term by term.
Right working
$sin^{3} A + cos^{3} A = (sin A + cos A) (1 - sin A cos A)$ .
Do this: Look for $a^{3} \pm b^{3}$ , $a^{2} - b^{2}$ and $(a \pm b)^{2}$ patterns.
Usually costs: Reported in 2024 Q3(i) and 2025 Q4(iii).

Exam tips

2025 Q4(iii) (4 marks): write $cot A + tan A - 1$ over $sin A cos A$ , recognise $sin^{3} A + cos^{3} A$ in the numerator, cancel, and reach $sec A cosec A$ .
2024 Q3(i) (4 marks): factorise $sin^{3} θ + cos^{3} θ$ first, then the identity follows in two lines.
Write the identity used in brackets beside the step where you use it.
Check with $A = 3 0^{\circ}$ in rough work before you start; it catches misread questions.

Practice questions on Proving 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
$\frac{t a n A}{s e c A}$ equals:
(a) $sin A$
(b) $cos A$
(c) $cot A$
(d) $cosec A$
Hint
Convert to sin and cos.
Answer
(a)
Q2 · 2 marks · Short answerBoard level
Prove that $sec^{2} A + cosec^{2} A = sec^{2} A cosec^{2} A$ .
Hint
Common denominator $sin^{2} A cos^{2} A$ .
Answer
LHS $= \frac{1}{c o s ^{2} A} + \frac{1}{s i n ^{2} A} = \frac{s i n ^{2} A + c o s ^{2} A}{s i n ^{2} A c o s ^{2} A} = \frac{1}{s i n ^{2} A c o s ^{2} A} =$ RHS.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $\frac{1 - c o s ^{2} A}{c o s ^{2} A} = tan^{2} A$ .
Reason (R): $1 - cos^{2} A = sin^{2} A$ .
(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
Substitute (R) into (A).
Answer
(c)
Q4 · 3 marks · Short answerHarder board level
Prove that $\frac{t a n A}{s e c A - 1} + \frac{t a n A}{s e c A + 1} = 2 cosec A$ .
Hint
Common denominator $sec^{2} A - 1 = tan^{2} A$ .
Answer
LHS $= \frac{t a n A \cdot 2 s e c A}{s e c ^{2} A - 1} = \frac{2 s e c A}{t a n A} = \frac{2}{c o s A} \times \frac{c o s A}{s i n A} = 2 cosec A$ .
Q5 · 3 marks · Short answerHarder board level
Prove that $sec^{2} A + cosec^{2} A = tan A + cot A$ for $0^{\circ} < A < 9 0^{\circ}$ .
Hint
$tan A cot A = 1$ .
Answer
$sec^{2} A + cosec^{2} A = (1 + tan^{2} A) + (1 + cot^{2} A) = tan^{2} A + 2 tan A cot A + cot^{2} A = (tan A + cot A)^{2}$ . Both are positive, so the square root is $tan A + cot A$ .
Q6 · 3 marks · Short answerHarder board level
Prove that $\frac{c o t A - c o s A}{c o t A + c o s A} = \frac{cosec A - 1}{cosec A + 1}$ .
Hint
Divide the numerator and denominator by $cos A$ .
Answer
LHS $= \frac{\frac{1}{s i n A} - 1}{\frac{1}{s i n A} + 1} = \frac{cosec A - 1}{cosec A + 1} =$ RHS.
Q7 · 4 marks · Long answerHOTS / competency
Prove that $\frac{s i n A}{c o t A + cosec A} = 2 + \frac{s i n A}{c o t A - cosec A}$ .
Hint
Show that LHS minus the fraction on the right equals 2.
Answer
$\frac{s i n A}{c o t A + cosec A} - \frac{s i n A}{c o t A - cosec A} = sin A \cdot \frac{( c o t A - cosec A ) - ( c o t A + cosec A )}{c o t ^{2} A - cosec ^{2} A} = sin A \cdot \frac{- 2 cosec A}{- 1} = 2$ . Rearranging gives the identity.

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. $\frac{1}{s e c A - t a n A}$ equals?Show answer
$sec A + tan A$
2. $tan A + cot A$ in terms of $sin A$ , $cos A$ ?Show answer
$\frac{1}{s i n A c o s A}$
3.Three ways of writing 1?Show answer
$sin^{2} A + cos^{2} A$ , $sec^{2} A - tan^{2} A$ , $cosec^{2} A - cot^{2} A$
4.Conjugate of $1 - sin A$ ?Show answer
$1 + sin A$
5.Which side do you start from?Show answer
The more complicated side.

Proving Trigonometric Identities: frequently asked questions

What is the best way to start proving a trigonometric identity?

Start with the more complicated side. If it has many different ratios, convert everything to sine and cosine; if it has fractions, combine them; if a denominator is 1 ± sin A or sec A ± tan A, multiply by the conjugate.

Can I work on both sides of an identity?

Yes, if you simplify each side separately to the same expression and then write LHS = RHS. Do not move terms from one side to the other as in an equation.

Which identities are in the ICSE Class 10 syllabus?

sin²A + cos²A = 1, 1 + tan²A = sec²A and 1 + cot²A = cosec²A, for 0° ≤ A ≤ 90°, used to prove and simplify expressions.