Trigonometric Identities ICSE Class 10: The Three Identities and How to Prove Results

What Trigonometric Identities is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Mathematics. Trigonometric Identities is in the 5. Trigonometry (a) part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Mathematics syllabus, examination year 2027

Key concepts

Ratios, reciprocals and quotients

In a right triangle with angle $A$: $\sin A=\frac{\text{opp}}{\text{hyp}}$, $\cos A=\frac{\text{adj}}{\text{hyp}}$, $\tan A=\frac{\text{opp}}{\text{adj}}$.

Reciprocals: $\mathrm{cosec}A=\frac{1}{\sin A}$, $\sec A=\frac{1}{\cos A}$, $\cot A=\frac{1}{\tan A}$.
Quotients: $\tan A=\frac{\sin A}{\cos A}$, $\cot A=\frac{\cos A}{\sin A}$.

These, with the three identities, are all you need.

Worked example

If $\sin A=\frac{3}{5}$ and $A$ is acute, find $\cos A$, $\tan A$ and $\sec A$.

Try it first, then show the solutionSolution

1. Sides 3 (opposite) and 5 (hypotenuse), so adjacent $=\sqrt{25-9}=4$.
2. $\cos A=\frac{4}{5}$, $\tan A=\frac{3}{4}$, $\sec A=\frac{5}{4}$.

Answer: cos A = 4/5, tan A = 3/4, sec A = 5/4

Identity 1: sin²A + cos²A = 1

Divide Pythagoras ${\text{opp}}^2+{\text{adj}}^2={\text{hyp}}^2$ by ${\text{hyp}}^2$:
${\sin }^{2}A+{\cos }^{2}A=1.$
Useful rearrangements: $1-{\sin }^{2}A={\cos }^{2}A$ and $1-{\cos }^{2}A={\sin }^{2}A$; also $(1-\sin A)(1+\sin A)={\cos }^{2}A$.

Note ${\sin }^{2}A$ means $(\sin A{)}^2$, not $\sin (A^2)$.

Worked example

Prove that $(1-{\cos }^{2}A){\mathrm{cosec}}^{2}A=1$.

Try it first, then show the solutionSolution

1. LHS $={\sin }^{2}A\times \frac{1}{{\sin }^{2}A}$.
2. $=1=$ RHS.

Answer: Proved

Identity 2: 1 + tan²A = sec²A

Divide Pythagoras by ${\text{adj}}^2$:
$1+{\tan }^{2}A={\sec }^{2}A,{\sec }^{2}A-{\tan }^{2}A=1.$
Factorised: $(\sec A-\tan A)(\sec A+\tan A)=1$, so $\sec A-\tan A$ and $\sec A+\tan A$ are reciprocals. This is a quick route to $\sec A$ and $\tan A$ from their sum.

Worked example

If $\sec A+\tan A=3$, find $\sec A$ and $\tan A$.

Try it first, then show the solutionSolution

1. $\sec A-\tan A=\frac{1}{3}$ (reciprocal).
2. Adding: $2\sec A=\frac{10}{3}$, so $\sec A=\frac{5}{3}$.
3. Subtracting: $2\tan A=\frac{8}{3}$, so $\tan A=\frac{4}{3}$.

Answer: sec A = 5/3, tan A = 4/3

Identity 3: 1 + cot²A = cosec²A

Divide Pythagoras by ${\text{opp}}^2$:
$1+{\cot }^{2}A={\mathrm{cosec}}^{2}A,{\mathrm{cosec}}^{2}A-{\cot }^{2}A=1.$
So $(\mathrm{cosec}A-\cot A)(\mathrm{cosec}A+\cot A)=1$.

The minus sign matters: ${\mathrm{cosec}}^{2}A+{\cot }^{2}A$ is not 1.

Worked example

Prove that ${\cot }^{2}A-{\cos }^{2}A={\cot }^{2}A{\cos }^{2}A$.

Try it first, then show the solutionSolution

1. LHS $=\frac{{\cos }^{2}A}{{\sin }^{2}A}-{\cos }^{2}A={\cos }^{2}A\left(\frac{1}{{\sin }^{2}A}-1\right)$.
2. $={\cos }^{2}A({\mathrm{cosec}}^{2}A-1)={\cos }^{2}A{\cot }^{2}A=$ RHS.

Answer: Proved

How to prove an identity

1. Start from the more complicated side and work towards the other. Do not move terms across the equals sign.
2. If stuck, write everything in sin and cos.
3. Combine fractions over a common denominator; simplify the numerator using an identity.
4. Look for conjugates: multiply $1-\sin A$ by $1+\sin A$ to get ${\cos }^{2}A$.
5. Use algebraic identities: $(a\pm b{)}^2$, $a^2-b^2$, $a^3\pm b^3$.
6. End with 'LHS = RHS' or '= RHS'.

Worked example

Prove that $(\sin \theta +\cos \theta {)}^2+(\sin \theta -\cos \theta {)}^2=2$.

Try it first, then show the solutionSolution

1. LHS $=({\sin }^2\theta +2\sin \theta \cos \theta +{\cos }^2\theta )+({\sin }^2\theta -2\sin \theta \cos \theta +{\cos }^2\theta )$.
2. $=2({\sin }^2\theta +{\cos }^2\theta )=2=$ RHS.

Answer: Proved

Algebraic identities inside trigonometry

Treat $\sin A$ and $\cos A$ as $a$ and $b$:

• $a^3+b^3=(a+b)(a^2-ab+b^2)$, so ${\sin }^{3}A+{\cos }^{3}A=(\sin A+\cos A)(1-\sin A\cos A)$.
• $a^3-b^3=(a-b)(a^2+ab+b^2)$, so ${\sin }^{3}A-{\cos }^{3}A=(\sin A-\cos A)(1+\sin A\cos A)$.
• $a^4-b^4=(a^2-b^2)(a^2+b^2)$, so ${\sin }^{4}A-{\cos }^{4}A={\sin }^{2}A-{\cos }^{2}A$.

In each case, $a^2+b^2$ becomes 1.

Worked example

Prove that $\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta -\cos \theta }=1+\sin \theta \cos \theta$ (for $\sin \theta \ne \cos \theta$).

Try it first, then show the solutionSolution

1. Numerator $=(\sin \theta -\cos \theta )({\sin }^2\theta +\sin \theta \cos \theta +{\cos }^2\theta )$.
2. Cancel $(\sin \theta -\cos \theta )$: LHS $=1+\sin \theta \cos \theta =$ RHS.

Answer: Proved

Finding values and eliminating the angle

Identities also let you:

• find all ratios from one ratio (draw the triangle, or use an identity);
• eliminate $\theta$: if $x=a\sec \theta$ and $y=b\tan \theta$, then $\frac{x^2}{a^2}-\frac{y^2}{b^2}={\sec }^2\theta -{\tan }^2\theta =1$.

A quick numerical check (put $A=30^{\circ }$, $45^{\circ }$ or $60^{\circ }$ into both sides) catches errors before you write a proof, but a numerical check is not a proof.

Worked example

If $x=2\cos \theta$ and $y=2\sin \theta$, find a relation between $x$ and $y$ without $\theta$.

Try it first, then show the solutionSolution

1. $\cos \theta =\frac{x}{2}$, $\sin \theta =\frac{y}{2}$.
2. $\frac{x^2}{4}+\frac{y^2}{4}=1$, so $x^2+y^2=4$.

Answer: x² + y² = 4

Maths in real life: Trigonometric Identities

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.

Topics in this chapter

• Proving Trigonometric Identities
• The Three Fundamental Trigonometric Identities
• Algebraic Identities in Trigonometry

Trigonometric Identities formula sheet

Identity 1

${\sin }^{2}A+{\cos }^{2}A=1$

1 − sin²A = cos²A; 1 − cos²A = sin²A.

Identity 2

$1+{\tan }^{2}A={\sec }^{2}A$

(sec A − tan A)(sec A + tan A) = 1.

Identity 3

$1+{\cot }^{2}A={\mathrm{cosec}}^{2}A$

cosec²A − cot²A = 1 (minus, not plus).

Quotients

$\tan A=\frac{\sin A}{\cos A},\cot A=\frac{\cos A}{\sin A}$

Reciprocals: cosec = 1/sin, sec = 1/cos.

Sum of cubes

${\sin }^{3}A+{\cos }^{3}A=(\sin A+\cos A)(1-\sin A\cos A)$

Difference: (sin A − cos A)(1 + sin A cos A).

Standard values

$\sin 30^{\circ }=\frac{1}{2},\sin 45^{\circ }=\frac{1}{\sqrt{2}},\sin 60^{\circ }=\frac{\sqrt{3}}{2}$

cos 30° = √3/2, cos 60° = 1/2; tan 45° = 1.

How the board asks Trigonometric Identities

The question patterns to expect, and the method that works for each.

Tricks to remember Trigonometric Identities

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

• One square identity, divided three ways

  Pythagoras ÷ hyp² gives sin² + cos² = 1; ÷ adj² gives 1 + tan² = sec²; ÷ opp² gives 1 + cot² = cosec².

  Use it when: Rebuilding the identities if you forget.

• Co-functions pair up

  tan goes with sec; cot goes with cosec ('the co's stay together').

  Use it when: Choosing identity 2 or 3.

• Minus makes one

  sec² − tan² = 1 and cosec² − cot² = 1: the bigger one minus the smaller one gives 1.

  Use it when: Avoiding the cosec² + cot² trap.

• Conjugate trick

  Multiply 1 − sin A by 1 + sin A to get cos²A (and similarly for cos, sec–tan, cosec–cot).

  Use it when: Square-root and fraction proofs.

Trigonometric Identities: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Trigonometric Identities

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

• ✗Mistake: Wrong sign in identity 3

  CISCE examiners' comments, 2024: in a statement MCQ, candidates were not familiar with cosec²θ − cot²θ = 1 and accepted cosec²θ + cot²θ = 1 as valid.

  Wrong working

  cosec²θ + cot²θ = 1

  Right working

  cosec²θ − cot²θ = 1, i.e. 1 + cot²θ = cosec²θ

  ✓Do this: Rebuild it from Pythagoras divided by opp².

  Usually costs: 1 mark.

• ✗Mistake: Not using a³ + b³

  CISCE examiners' comments, 2024 and 2025: candidates failed to apply a³ + b³ = (a + b)(a² − ab + b²) to factorise sin³θ + cos³θ, and so could not finish the proof.

  ✓Do this: When you see cubes of sin and cos, factorise first; a² + b² becomes 1.

  Usually costs: 2 to 3 marks in a 4-mark proof.

• ✗Mistake: Errors with a common denominator

  CISCE examiners' comments, 2024 and 2025: candidates who took a common denominator (such as sin A cos A) made simplification errors.

  ✓Do this: Write each step on a new line; expand brackets fully before cancelling.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Not linking the assertion to the reason

  CISCE examiners' comments, 2025: in an assertion–reason question on sin²A + sin A = 1, candidates could not see that the reason (1 − sin²A = cos²A) was the step that proves the assertion.

  ✓Do this: Try to prove (A) using (R). If (R) is the key step, it is the correct explanation.

  Usually costs: 1 mark.

• ✗Mistake: Working on both sides together

  A proof that rearranges the identity being proved, or moves terms across the equals sign, assumes what it is meant to show.

  ✓Do this: Transform one side until it becomes the other; write '= RHS' at the end.

  Usually costs: Method marks in a proof.

Practice questions on Trigonometric Identities

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

1. Q1 · 1 mark · MCQBoard level

   The value of ${\mathrm{cosec}}^2\theta -{\cot }^2\theta$ is:
   (a) 0
   (b) 1
   (c) $-1$
   (d) 2

   Hint

   Identity 3.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   $(1-{\sin }^{2}A){\sec }^{2}A$ equals:
   (a) 0
   (b) 1
   (c) ${\tan }^{2}A$
   (d) ${\cos }^{2}A$

   Hint

   cos²A × 1/cos²A.

   Answer

   (b)

3. Q3 · 1 mark · MCQHarder board level

   If $\tan A=\frac{4}{3}$ and $A$ is acute, then $\sin A+\cos A$ is:
   (a) $\frac{7}{3}$
   (b) $\frac{7}{5}$
   (c) $\frac{1}{5}$
   (d) $\frac{12}{25}$

   Hint

   3-4-5 triangle.

   Answer

   (b)

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): ${\sin }^{4}A-{\cos }^{4}A={\sin }^{2}A-{\cos }^{2}A$.
   Reason (R): $a^2-b^2=(a-b)(a+b)$, and ${\sin }^{2}A+{\cos }^{2}A=1$.
   (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

   Factorise the fourth powers.

   Answer

   (c)

5. Q5 · 4 marks · Short answerHarder board level

   Prove that $(\mathrm{cosec}A-\cot A{)}^2=\frac{1-\cos A}{1+\cos A}$.

   Hint

   Write in sin and cos; sin²A = (1 − cos A)(1 + cos A).

   Answer

   LHS $={\left(\frac{1-\cos A}{\sin A}\right)}^2=\frac{(1-\cos A{)}^2}{1-{\cos }^{2}A}=\frac{(1-\cos A{)}^2}{(1-\cos A)(1+\cos A)}=\frac{1-\cos A}{1+\cos A}$.

6. Q6 · 4 marks · Short answerHarder board level

   If $\sec A+\tan A=p$, show that $\sin A=\frac{p^2-1}{p^2+1}$. Check with $p=3$.

   Hint

   Find sec A and tan A in terms of p.

   Answer

   $\sec A-\tan A=\frac{1}{p}$, so $\sec A=\frac{p^2+1}{2p}$, $\tan A=\frac{p^2-1}{2p}$. $\sin A=\frac{\tan A}{\sec A}=\frac{p^2-1}{p^2+1}$. With $p=3$: $\frac{8}{10}=\frac{4}{5}$, matching $\sec A=\frac{5}{3}$, $\tan A=\frac{4}{3}$.

7. Q7 · 3 marks · Long answerBoard level

   Prove that $\tan A+\cot A=\sec A\mathrm{cosec}A$.

   Hint

   Common denominator sin A cos A.

   Answer

   LHS $=\frac{{\sin }^{2}A+{\cos }^{2}A}{\sin A\cos A}=\frac{1}{\sin A\cos A}=\sec A\mathrm{cosec}A$.

8. Q8 · 4 marks · Long answerHarder board level

   Prove that $(1+\cot A-\mathrm{cosec}A)(1+\tan A+\sec A)=2$.

   Hint

   Convert to sin and cos; use (a + b)(a − b).

   Answer

   LHS $=\frac{\sin A+\cos A-1}{\sin A}\times \frac{\cos A+\sin A+1}{\cos A}=\frac{(\sin A+\cos A{)}^2-1}{\sin A\cos A}=\frac{2\sin A\cos A}{\sin A\cos A}=2$.

9. Q9 · 4 marks · Long answerHOTS / competency

   If $\tan A+\sin A=m$ and $\tan A-\sin A=n$, show that $m^2-n^2=4\sqrt{mn}$.

   Hint

   Find m² − n² and mn separately.

   Answer

   $m^2-n^2=(m+n)(m-n)=4\tan A\sin A$. $mn={\tan }^{2}A-{\sin }^{2}A={\sin }^{2}A\left(\frac{1}{{\cos }^{2}A}-1\right)={\sin }^{2}A{\tan }^{2}A$, so $4\sqrt{mn}=4\tan A\sin A$ (both positive for acute $A$). Equal.

10. Q10 · 4 marks · Long answerHarder board level

    Prove that $\frac{\sec A-1}{\sec A+1}=\frac{1-\cos A}{1+\cos A}$, and check it for $A=60^{\circ }$.

    Hint

    Multiply top and bottom by cos A.

    Answer

    Multiply numerator and denominator by $\cos A$: $\frac{1-\cos A}{1+\cos A}$. At $60^{\circ }$: $\frac{2-1}{2+1}=\frac{1}{3}$ and $\frac{0.5}{1.5}=\frac{1}{3}$ ✓

11. Q11 · 1 mark · MCQHOTS / competency

    If $\sin A=\cos A$ with $0^{\circ }\le A\le 90^{\circ }$, then ${\sin }^{4}A+{\cos }^{4}A$ equals:
    (a) 1
    (b) $\frac{1}{2}$
    (c) $\frac{1}{4}$
    (d) 2

    Hint

    sin²A = cos²A = 1/2.

    Answer

    (b)

12. Q12 · 5 marks · Long answerHOTS / competency

    (a) Prove that $(\sin A+\cos A)(\tan A+\cot A)=\sec A+\mathrm{cosec}A$. (b) Verify it for $A=45^{\circ }$. (c) Hence find its value for $A=30^{\circ }$, correct to 2 decimal places.

    Hint

    tan A + cot A = 1/(sin A cos A).

    Answer

    (a) $(\sin A+\cos A)\times \frac{1}{\sin A\cos A}=\frac{1}{\cos A}+\frac{1}{\sin A}=\sec A+\mathrm{cosec}A$. (b) LHS $=\sqrt{2}\times 2=2\sqrt{2}$; RHS $=\sqrt{2}+\sqrt{2}=2\sqrt{2}$ ✓. (c) $\sec 30^{\circ }+\mathrm{cosec}30^{\circ }=\frac{2}{\sqrt{3}}+2\approx 3.15$.

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. 1.Identity linking sin and cos?Show answer

   sin²A + cos²A = 1

2. 2.Identity linking tan and sec?Show answer

   1 + tan²A = sec²A

3. 3.Identity linking cot and cosec?Show answer

   1 + cot²A = cosec²A

4. 4.(sec A − tan A)(sec A + tan A) = ?Show answer

   1

5. 5.Factorise sin³A + cos³A.Show answer

   (sin A + cos A)(1 − sin A cos A)

6. 6.1 − sin²A = ?Show answer

   cos²A

7. 7.Range of A for the ICSE identities?Show answer

   0° ≤ A ≤ 90°

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you derive all three identities?
• ☐Do you remember cosec² − cot² = 1 (minus)?
• ☐Do you start from the more complicated side?
• ☐Can you factorise sin³ ± cos³?
• ☐Can you use sec + tan and sec − tan as reciprocals?
• ☐Do you end every proof with '= RHS'?

Trigonometric Identities: frequently asked questions