Introduction to Trigonometry Class 10 Notes PDF: Formulas, Identities and Questions

Introduction to Trigonometry Class 10 notes explain how the angles and sides of a right-angled triangle are connected through six trigonometric ratios. This guide covers the complete NCERT Class 10 chapter: side identification, sin–cos–tan formulas, standard values, complementary angles, identities, worked examples, common mistakes, competency-based questions and a printable revision sheet.

The current CBSE Class 10 curriculum places Introduction to Trigonometry, trigonometric identities, and heights and distances within a 12-mark Trigonometry unit. That is a unit-level allocation: it does not mean Chapter 8 alone will always carry 12 marks.

Introduction to Trigonometry Chapter 8: Overview

Introduction to Trigonometry studies fixed ratios between the sides of a right-angled triangle for a chosen acute angle.

By the end of the chapter, you should be able to:

1. Identify the opposite, adjacent and hypotenuse sides.
2. Define and calculate all six trigonometric ratios.
3. Find unknown ratios when one ratio is given.
4. recall or derive standard values for 0°, 30°, 45°, 60° and 90°.
5. Use relationships between trigonometric ratios.
6. Prove simple trigonometric identities.
7. Explain why some ratios are undefined at 0° or 90°.

The official CBSE syllabus includes ratios of an acute angle, proof that the ratios are well-defined, standard-angle values, relationships between ratios and simple identities derived from sin²A + cos²A = 1.

Download Introduction to Trigonometry Class 10 Notes PDF

The downloadable Introduction to Trigonometry Class 10 Notes PDF should provide the complete lesson, not merely a short description of what the notes contain.

What is trigonometry?

Trigonometry is the study of relationships between the angles and sides of triangles. In Class 10, the focus is mainly on right-angled triangles and their acute angles. NCERT introduces trigonometry through situations involving heights and distances before defining the ratios used in this chapter.

The word is formed from Greek roots associated with triangle and measurement.

What belongs in Chapter 8?

Chapter 8 focuses on:

• Trigonometric ratios
• Standard-angle values
• Relationships between ratios
• Complementary angles
• Trigonometric identities
• Simple identity proofs

What does not belong in the core Chapter 8 notes?

The following are applications of trigonometry and are treated separately in the curriculum:

• Line of sight
• Angle of elevation
• Angle of depression
• Finding the height of buildings or towers
• Finding horizontal distances from observation points

CBSE lists Introduction to Trigonometry, identities, and heights and distances as three separate components within Unit V.

Also Check: Applications of Trigonometry notes 

Parts of a Right-Angled Triangle

A right-angled triangle has one 90° angle, one hypotenuse and two other sides whose names depend on the selected reference angle.

Right angle

Right angle — The angle measuring exactly 90°.

A small square is normally drawn at the right angle.

Hypotenuse

Hypotenuse — The side directly opposite the 90° angle.

The hypotenuse:

• Is always the longest side.
• Does not change when you select a different acute angle.
• Is used in the definitions of sine, cosine, cosecant and secant.

Reference angle

Reference angle — The acute angle from which the triangle’s other sides are being described.

Before choosing a trigonometric ratio, mark the reference angle clearly.

Opposite side

Opposite side — The side directly across from the reference angle.

The opposite side does not touch the selected angle.

Adjacent side

Adjacent side — The non-hypotenuse side that touches the reference angle.

Do not call the hypotenuse the adjacent side even though it also touches the angle.

Why opposite and adjacent can switch

Opposite and adjacent are relative names, so they switch when the reference angle changes.

Suppose triangle ABC is right-angled at B:

• For angle A, side BC is opposite and AB is adjacent.
• For angle C, side AB is opposite and BC is adjacent.
• Side AC remains the hypotenuse in both cases.

NCERT explicitly shows that the side positions change when angle C is considered instead of angle A.

Three-step side-identification method

1. Find the right angle. The side opposite it is the hypotenuse.
2. Circle the reference angle.
3. Label the remaining sides. The side across from the reference angle is opposite; the other non-hypotenuse side is adjacent.

Quick check: The hypotenuse never changes, but the opposite and adjacent sides may change.

The Six Trigonometric Ratios

The six trigonometric ratios are sine, cosine, tangent, cosecant, secant and cotangent.

Let:

• P = perpendicular or opposite side
• B = base or adjacent side
• H = hypotenuse

These definitions match the six ratios given in the official NCERT chapter.

Sine

Sine compares the side opposite the angle with the hypotenuse.

sin A = opposite / hypotenuse

Cosine

Cosine compares the side adjacent to the angle with the hypotenuse.

cos A = adjacent / hypotenuse

Tangent

Tangent compares the side opposite the angle with the adjacent side.

tan A = opposite / adjacent

Cosecant

Cosecant is the reciprocal of sine.

cosec A = hypotenuse / opposite = 1 / sin A

Secant

Secant is the reciprocal of cosine.

sec A = hypotenuse / adjacent = 1 / cos A

Cotangent

Cotangent is the reciprocal of tangent.

cot A = adjacent / opposite = 1 / tan A

Memory aid

For the three main ratios, many students use:

• SOH: Sine = Opposite / Hypotenuse
• CAH: Cosine = Adjacent / Hypotenuse
• TOA: Tangent = Opposite / Adjacent

A memory aid is useful for revision, but always identify the reference angle and triangle sides before applying it.

Worked example: find all six ratios

In right triangle ABC, right-angled at B:

• Opposite to angle A = 6 cm
• Adjacent to angle A = 8 cm
• Hypotenuse = 10 cm

Therefore:

• sin A = 6/10 = 3/5
• cos A = 8/10 = 4/5
• tan A = 6/8 = 3/4
• cosec A = 10/6 = 5/3
• sec A = 10/8 = 5/4
• cot A = 8/6 = 4/3

Why Trigonometric Ratios Stay the Same

Trigonometric ratios stay constant for a fixed angle because right triangles containing that angle are similar.

Consider three right triangles with proportional side lengths:

The side lengths increase, but each ratio remains unchanged.

NCERT proves this using the AA similarity criterion: triangles containing the same acute angle have proportional corresponding sides, so their trigonometric ratios are equal.

How to Choose the Correct Trigonometric Ratio

Choose a trigonometric ratio by identifying the reference angle, the known side and the side you need to find.

Four-step ratio-selection method

1. Mark the reference angle.
2. Label opposite, adjacent and hypotenuse.
3. Identify the known and unknown sides.
4. Select the ratio containing only those two sides.

Example 1: opposite and hypotenuse

A right triangle has:

• Hypotenuse = 12 cm
• Angle = 30°
• Opposite side = x

Use sine:

sin 30° = x/12

Since sin 30° = 1/2:

1/2 = x/12

x = 6 cm

Example 2: adjacent and hypotenuse

A right triangle has:

• Hypotenuse = 10 cm
• Adjacent side = 5√3 cm
• Reference angle = A

Use cosine:

cos A = 5√3/10 = √3/2

Therefore, A = 30°.

Example 3: opposite and adjacent

A right triangle has:

• Opposite side = 8 cm
• Adjacent side = 8 cm

Use tangent:

tan A = 8/8 = 1

Therefore, A = 45°.

When should reciprocal ratios be used?

Reciprocal ratios are useful when the question directly gives or asks for cosec, sec or cot, but most side-finding questions can first be approached with sin, cos or tan.

For example:

sec A = hypotenuse / adjacent

If sec A = 5/4, then:

cos A = 4/5

Relationships Between Trigonometric Ratios

The six ratios are connected by reciprocal and quotient relationships.

Reciprocal relationships

Quotient relationships

tan A = sin A / cos A

cot A = cos A / sin A

These relationships follow directly from the side definitions. For example:

sin A / cos A

= (opposite/hypotenuse) ÷ (adjacent/hypotenuse)

= opposite/adjacent

= tan A

Important notation: reciprocal is not inverse

cosec A is the reciprocal of sin A, but sin⁻¹x represents inverse sine and is not the same expression.

• cosec A = 1/sin A
• sin⁻¹x means the angle whose sine is x

NCERT specifically warns that (sin A)⁻¹ as a reciprocal can represent cosecant, while sin⁻¹A has a different meaning studied in higher classes.

Important notation: sin²A

sin²A means (sin A)², not sin(A²).

Similarly:

• cos²A = (cos A)²
• tan²A = (tan A)²

Standard Trigonometric Values for 0°, 30°, 45°, 60° and 90°

The standard trigonometric table gives exact values of all six ratios at 0°, 30°, 45°, 60° and 90°.

The official NCERT table also notes that as the angle increases from 0° to 90°, sine increases from 0 to 1, while cosine decreases from 1 to 0.

[IMAGE PLACEHOLDER: Colour-coded standard trigonometric values table optimized for mobile and print. Undefined cells should use a clear symbol and not display zero. Alt text: “Trigonometric values of 0, 30, 45, 60 and 90 degrees.”]

How to derive the values for 45°

The 45° values come from an isosceles right triangle with equal perpendicular sides.

Let both shorter sides equal 1.

By the Pythagoras theorem:

hypotenuse² = 1² + 1² = 2

hypotenuse = √2

Therefore:

• sin 45° = 1/√2
• cos 45° = 1/√2
• tan 45° = 1

How to derive the values for 30° and 60°

The 30° and 60° values come from dividing an equilateral triangle into two congruent right triangles.

Start with an equilateral triangle of side 2.

Dividing it vertically creates:

• Hypotenuse = 2
• Shorter side = 1
• Other side = √3

For the 30° angle:

• sin 30° = 1/2
• cos 30° = √3/2
• tan 30° = 1/√3

For the 60° angle:

• sin 60° = √3/2
• cos 60° = 1/2
• tan 60° = √3

NCERT uses the same equal-sided and equilateral-triangle constructions to derive these values.

How to remember sine values

Use the sequence:

sin 0° = √0/2

sin 30° = √1/2

sin 45° = √2/2

sin 60° = √3/2

sin 90° = √4/2

The cosine sequence is the sine sequence in reverse:

cos 0° = √4/2

cos 30° = √3/2

cos 45° = √2/2

cos 60° = √1/2

cos 90° = √0/2

Then calculate:

tan A = sin A / cos A

Why is tan 90° not defined?

tan 90° is not defined because tan A = sin A/cos A and cos 90° = 0.

Therefore:

tan 90° = 1/0

Division by zero is not defined.

Why are cosec 0° and cot 0° not defined?

cosec 0° = 1/sin 0° = 1/0

cot 0° = cos 0°/sin 0° = 1/0

Both expressions involve division by zero, so both are undefined.

Why is sec 90° not defined?

sec 90° = 1/cos 90° = 1/0

Therefore, sec 90° is undefined.

NCERT motivates the endpoint values by showing how the triangle changes as an acute angle approaches 0° or 90°.

Trigonometric Ratios of Complementary Angles

Two acute angles in a right triangle are complementary, so the sine of one equals the cosine of the other.

If:

A + B = 90°

then:

Why do complementary relationships work?

The opposite side for one acute angle is the adjacent side for the other acute angle.

In the same right triangle:

• What is opposite to angle A is adjacent to angle B.
• What is adjacent to angle A is opposite to angle B.
• The hypotenuse remains the same.

Therefore:

sin A = opposite to A / hypotenuse

cos B = adjacent to B / hypotenuse

Because those numerator sides are the same:

sin A = cos B

Since B = 90° − A:

sin A = cos(90° − A)

Solved example 1

Find:

`sin 35° − cos .”]

Solved example 1

Find:

sin 55°

Because:

cos 55° = cos(90° − 35°) = sin 35°

Therefore:

sin 35° − cos 55° = 0

Solved example 2

If:

tan A = cot 40°

then:

tan A = tan(90° − 40°)

tan A = tan 50°

For acute angles:

A = 50°

Trigonometric Identities Class 10

A trigonometric identity is an equality involving trigonometric ratios that is true for every permitted value of the angle.

Identity versus equation

The three main trigonometric identities

1. sin²A + cos²A = 1
2. 1 + tan²A = sec²A
3. 1 + cot²A = cosec²A

CBSE specifies the proof and application of sin²A + cos²A = 1 and lmple identities. citeturn796718view2

Derivation of sin²A + cos²A = 1

In right triangle ABC, suppose:

• Opposite side = BC
• Adjacent side = AB
• Hypotenuse = AC

By the Pythagoras theorem:

AB² + BC² = AC²

Divide every term by AC²:

AB²/AC² + BC²/AC² = AC²/AC²

(AB/AC)² + (BC/AC)² = 1

cos²A + sin²A = 1

Therefore:

sin²A + cos²A = 1

Derivation of 1 + tan²A = sec²A

Start with:

AB² + BC² = AC²

Divide by AB²:

1 + BC²/AB² = AC²/AB²

1 + tan²A = sec²A

Derivation of 1 + cot²A = cosec²A

Start with:

AB² + BC² = AC²

Divide by BC²:

AB²/BC² + 1 = AC²/BC²

cot²A + 1 = cosec²A

NCERT derives all three identities fr citeturn647573view5turn647573view6

Equivalent forms to recognise

From sin²A + cos²A = 1:

• sin²A = 1 − cos²A
• cos²A = 1 − sin²A

From 1 + tan²A = sec²A:

• tan²A = sec²A − 1
• sec²A − tan²A = 1

From 1 + cot²A = cosec²A:

• cot²A = cosec²A − 1
• cosec²A − cot²A = 1

How to Prove Trigonometric Identities Step by Step

To prove a trigonometric identity, simplify the more complicated side using known identities and valid algebra until it becomes the other side.

Eight-step identity-proof method

1. Begin with the more complicated side.
2. Look for a recognisable identity.
3. Convert tan, cot, sec or cosec into sin and cos when useful.
4. Take a common denominator when fractions contain different denominators.
5. Factor before expanding whenever possible.
6. Use conjugates when an expression contains 1 ± sin A or 1 ± cos A.
7. Cancel factors only—not separate terms joined by addition or subtraction.
8. Stop when the expression becomes the target side.

When should everything be converted into sine and cosine?

Convert to sine and cosine when:

• An expression contains several different ratios.
• Reciprocal terms such as sec and cosec appear.
• Quotient terms such as tan and cot appear.
• No direct Pythagorean identity is visible.

Use:

• tan A = sin A/cos A
• cot A = cos A/sin A
• sec A = 1/cos A
• cosec A = 1/sin A

Worked proof 1: direct identity

Prove:

(1 − sin²A)/cos²A = 1

LHS

= (1 − sin²A)/cos²A

Using 1 − sin²A = cos²A:

= cos²A/cos²A

= 1

= RHS

Worked proof 2: converting secant and tangent

Prove:

(sec A − cos A)/tan A = sin A

LHS

= (1/cos A − cos A) ÷ (sin A/cos A)

= [(1 − cos²A)/cos A] × [cos A/sin A]

Using 1 − cos²A = sin²A:

= [sin²A/cos A] × [cos A/sin A]

= sin A

= RHS

Worked proof 3: common denominator

Prove:

1/(1 + sin A) + 1/(1 − sin A) = 2sec²A

LHS

= [(1 − sin A) + (1 + sin A)] / [(1 + sin A)(1 − sin A)]

= 2/(1 − sin²A)

Using 1 − sin²A = cos²A:

= 2/cos²A

= 2sec²A

= RHS

Worked proof 4: factorisation

Prove:

(1 − cos²A)/sin²A = 1

LHS

= (1 − cos²A)/sin²A

Using 1 − cos²A = sin²A:

= sin²A/sin²A

= 1

= RHS

Worked proof 5: conjugate method

Prove:

(1 − sin A)/cos A = cos A/(1 + sin A)

LHS

Multiply numerator and denominator by the conjugate 1 + sin A:

= [(1 − sin A)(1 + sin A)] / [cos A(1 + sin A)]

= (1 − sin²A) / [cos A(1 + sin A)]

Using 1 − sin²A = cos²A:

= cos²A / [cos A(1 + sin A)]

= cos A/(1 + sin A)

= RHS

How to present a proof clearly

Use the following board-style structure:

1. Write LHS =.
2. Show one valid change per line.
3. Name the identity when it is used.
4. Avoid beginning with both LHS and RHS and changing them randomly.
5. End with = RHS.

NCERT’s worked identity examples also simplify from the left-hand side through a citeturn647573view6turn647573view7

Common Mistakes in Class 10 Trigonometry

Most trigonometry errors come from incorrect side labels, confused notation, division by zero or invalid algebraic cancellation.

Invalid cancellation example

Incorrect:

(sin A + cos A)/sin A = cos A

You cannot cancel sin A from only one term in the numerator.

Correct:

(sin A + cos A)/sin A

= sin A/sin A + cos A/sin A

= 1 + cot A

Can tangent be greater than 1?

Yes.

For example:

tan 60° = √3

Since √3 ≈ 1.732, tangent can be greater than 1.

NCERT includes a true-or-false exercise specifically testing the misconceptions less than 1. citeturn796718view0

Solved Examples Based on Trigonometric Ratios

Worked examples show how to move from triangle information to ratios, standard values and identities.

Example 1: find the third side and all ratios

Level: Both Basic and Standard

In a right triangle, relative to angle A:

• Opposite = 5 cm
• Adjacent = 12 cm

Find the hypotenuse and all six ratios.

By the Pythagoras theorem:

hypotenuse² = 5² + 12²

= 25 + 144

= 169

hypotenuse = 13 cm

Therefore:

• sin A = 5/13
• cos A = 12/13
• tan A = 5/12
• cosec A = 13/5
• sec A = 13/12
• cot A = 12/5

Example 2: find the other ratios when tangent is given

Level: Both Basic and Standard

Given:

tan A = 3/4

Take:

• Opposite = 3k
• Adjacent = 4k

By the Pythagoras theorem:

hypotenuse = 5k

Therefore:

• sin A = 3/5
• cos A = 4/5
• cosec A = 5/3
• sec A = 5/4
• cot A = 4/3

Example 3: evaluate a standard-value expression

Level: Both Basic and Standard

Evaluate:

2sin 30° cos 60° + tan 45°

Substitute:

= 2(1/2)(1/2) + 1

= 1/2 + 1

= 3/2

Example 4: complementary angles

Level: Both Basic and Standard

Evaluate:

sin 25°/cos 65°

Since:

cos 65° = cos(90° − 25°) = sin 25°

Therefore:

sin 25°/cos 65° = 1

Example 5: competency-based scaling question

Level: Standard focus

Triangle 1 has sides 3, 4, 5. Triangle 2 has sides 12, 16, 20. Both contain the same acute angle opposite the shortest side.

Compare the sine of that angle.

Triangle 1:

sin A = 3/5

Triangle 2:

sin A = 12/20 = 3/5

The values are equal because the triangles are similar.

Example 6: error analysis

Level: Standard focus

A student writes:

sec²A − tan²A = sec²A − (sec²A + 1) = −1

Identify the error.

The identity is:

1 + tan²A = sec²A

Therefore:

tan²A = sec²A − 1

Correct calculation:

sec²A − tan²A

= sec²A − (sec²A − 1)

= 1

NCERT Exercise and Topic Map

The NCERT exercises progress from side ratios to standard-angle values and then to identity proofs.

Also Check: NCERT Class 10 Chapter 8 solutions 

Is NCERT enough for Class 10 trigonometry?

NCERT is the essential starting point, but students should also practise competency-based and timed questions after mastering the textbook exercises.

A useful order is:

1. Read the NCERT examples.
2. Solve every exercise question independently.
3. Check complete solutions.
4. Practise error-analysis and visual questions.
5. Complete a timed mixed test.

CBSE publishes curriculum-aligned competency-focused question resources designed to assess application

Mathematics Basic Versus Mathematics Standard

Both papers use the same core trigonometry concepts, but Mathematics Standard places more weight on application and higher-order reasoning.

These percentages describe the design of the complete Mathematics papers citeturn796718view3turn796718view4

Mathematics Basic practice

Focus on:

• Identifying sides correctly
• Direct ratio calculations
• Standard values
• Straightforward substitutions
• Simple identities
• Clear one- or two-step problems

Mathematics Standard practice

Add:

• Multi-step algebra
• Identity transformations
• Visual interpretation
• Error analysis
• Statement justification
• Similarity-based reasoning
• Questions combining more than one concept

Labels used in this guide

• Both: Suitable for Basic and Standard
• Basic focus: Direct concept or substitution
• Standard focus: Application or reasoning
• Challenge: Additional practice; check syllabus alignment before treating it as required

Introduction to Trigonometry Important Questions

Important trigonometry questions should cover definitions, standard values, ratio selection, identities, misconceptions and reasoning.

Multiple-choice questions

1. In a right triangle, the side opposite the right angle is called:

A. Adjacent
B. Perpendicular
C. Hypotenuse
D. Base

Answer: C. The hypotenuse is opposite the 90° angle.

2. If sin A = 3/5, then cosec A equals:

A. 3/5
B. 5/3
C. 4/5
D. 5/4

Answer: B. Cosecant is the reciprocal of sine.

3. tan 45° equals:

A. 0
B. 1/√3
C. 1
D. √3

Answer: C.

4. Which ratio is not defined at 90°?

A. Sine
B. Cosecant
C. Tangent
D. Cotangent

Answer: C. tan 90° = sin 90°/cos 90° = 1/0.

5. If A + B = 90°, then sin A equals:

A. sin B
B. cos B
C. tan B
D. sec B

Answer: B.

6. Which identity is correct?

A. sin²A − cos²A = 1
B. 1 − tan²A = sec²A
C. 1 + tan²A = sec²A
D. 1 + sec²A = tan²A

Answer: C.

7. If opposite = adjacent, then the acute angle is:

A. 0°
B. 30°
C. 45°
D. 60°

Answer: C, because tan A = 1.

8. sin⁻¹x and cosec x are:

A. Always equal
B. Both reciprocals
C. Different mathematical expressions
D. Equal only at 45°

Answer: C.

Short-answer questions

1. Define all six trigonometric ratios.
2. Explain why the hypotenuse does not depend on the reference angle.
3. Given cos A = 12/13, find the remaining five ratios.
4. Evaluate sin 60° cos 30° + cos 60° sin 30°.
5. Explain why sec 90° is not defined.
6. Prove sec²A − tan²A = 1.
7. Prove (1 − cos A)(1 + cos A) = sin²A.
8. If tan A = cot 25°, find acute angle A.
9. Explain why sine remains constant when a triangle is enlarged proportionally.
10. Identify and correct the error: sin A + cos A = 1.

Answers

3. If cos A = 12/13, use a 5–12–13 triangle:

• sin A = 5/13
• tan A = 5/12
• cosec A = 13/5
• sec A = 13/12
• cot A = 12/5

sin 60° cos 30° + cos 60° sin 30°

= (√3/2)(√3/2) + (1/2)(1/2)

= 3/4 + 1/4

= 1

5. sec 90° = 1/cos 90° = 1/0, which is undefined.
6. From 1 + tan²A = sec²A, rearrange to obtain sec²A − tan²A = 1.
7.  

(1 − cos A)(1 + cos A)

= 1 − cos²A

= sin²A

tan A = cot 25°

tan A = tan(90° − 25°)

tan A = tan 65°

Therefore, A = 65°.

9. Proportional enlargement creates a similar triangle, so corresponding side ratios remain equal.
10. sin A + cos A is not always 1. The correct identity is sin²A + cos²A = 1.

Class 10 Trigonometry Practice Test

This 20-mark test checks direct knowledge, application, identity proofs and competency-based reasoning.

Suggested time: 35 minutes
Calculator: Not required
Suitable for: Mathematics Standard; Questions 1–7 also suit Mathematics Basic

Section A: One mark each

1. Write cos A as a side ratio.
2. Find sin 30°.
3. State the reciprocal of tan A.
4. Is cosec 0° defined? Give a reason.
5. Complete: sec²A − tan²A = ____.

Section B: Two marks each

6. In a right triangle, opposite = 9 cm and adjacent = 12 cm. Find sin A and cos A.
7. Evaluate 2cos²45° + tan 30°cot 30°.
8. If sin A = 8/17, find cos A and tan A.

Section C: Three marks each

9. Prove:

(1 − sin²A)/cos²A = 1

10. A student says that tan A must always be less than 1 because it is a ratio. Explain why the statement is false and give an exact counterexample.

Solutions and marking points

1.

cos A = adjacent/hypotenuse

2.

sin 30° = 1/2

3.

cot A

4.

No. cosec 0° = 1/sin 0° = 1/0, which is undefined.

5.

1

6.

Hypotenuse:

√(9² + 12²) = √225 = 15

Therefore:

sin A = 9/15 = 3/5

cos A = 12/15 = 4/5

7.

2cos²45° + tan 30°cot 30°

= 2(1/√2)² + (1/√3)(√3)

= 2(1/2) + 1

= 2

8.

Given:

sin A = 8/17

Take:

• Opposite = 8
• Hypotenuse = 17

Adjacent:

= √(17² − 8²)

= √(289 − 64)

= √225

= 15

Therefore:

cos A = 15/17

tan A = 8/15

9.

LHS = (1 − sin²A)/cos²A

= cos²A/cos²A

= 1

= RHS

10.

Tangent compares the opposite and adjacent sides, so it can exceed 1 when the opposite side is longer.

For example:

tan 60° = √3 > 1

Class 10 Trigonometry Formula Sheet

A Class 10 trigonometry formula sheet should contain the six ratios, reciprocal relationships, One-Page Class 10 Trigonometry Formula Sheet

A Class standard values, complementary-angle rules and three identities.

Six ratios

• sin A = opposite/hypotenuse
• cos A = adjacent/hypotenuse
• tan A = opposite/adjacent
• cosec A = hypotenuse/opposite
• sec A = hypotenuse/adjacent
• cot A = adjacent/opposite

Reciprocal relationships

• cosec A = 1/sin A
• sec A = 1/cos A
• cot A = 1/tan A

Quotient relationships

• tan A = sin A/cos A
• cot A = cos A/sin A

Three identities

• sin²A + cos²A = 1
• 1 + tan²A = sec²A
• 1 + cot²A = cosec²A

Complementary relationships

• sin A = cos(90° − A)
• cos A = sin(90° − A)
• tan A = cot(90° − A)
• cot A = tan(90° − A)
• sec A = cosec(90° − A)
• cosec A = sec(90° − A)

Undefined values

• tan 90°
• sec 90°
• cosec 0°
• cot 0°

Proof checklist

1. Start with the more complicated side.
2. Look for a direct identity.
3. Convert to sine and cosine if helpful.
4. Take a common denominator.
5. Factor before cancelling.
6. Use a conjugate where useful.
7. Change one side systematically.
8. Stop when it becomes the other side.

How to Revise Introduction to Trigonometry

Effective trigonometry revision combines formula recall with side identification, worked problems and identity-proof practice.

Ten-minute one-shot revision

1. Two minutes: Write the six ratios from memory.
2. Two minutes: Reproduce the standard-value table.
3. Two minutes: Write the three identities.
4. Two minutes: Solve one ratio-selection question.
5. Two minutes: Complete one identity proof.

One-day revision plan

Session 1: concepts

• Identify opposite, adjacent and hypotenuse.
• Review all six ratios.
• Complete five direct ratio questions.

Session 2: standard values

• Derive 30°, 45° and 60°.
• Recall endpoint values.
• Complete five expression-evaluation questions.

Session 3: identities

• Derive the three identities.
• Complete three proofs using different first steps.
• Review common algebra mistakes.

Session 4: testing

• Complete the 20-mark practice test.
• Mark errors by concept.
• Repeat only the weak question types.

Seven-day practice plan

How much practice is enough?

There is no fixed number that guarantees a score. A student is ready when they can:

• Label sides without hesitation.
• Reproduce the standard table accurately.
• Select a ratio without guessing.
• Explain every step in an identity proof.
• Solve mixed questions under time pressure.
• Correct their own algebra mistakes.

Also Check: 

Class 10 Maths sample papers 

Class 10 trigonometry MCQs with answers