Introduction to Trigonometry – Class 10 Maths Chapter 8 Notes & NCERT PDF

Class 10MathematicsChapter 8NCERT book: Mathematics - Class 10

Notes and a simple summary of Chapter 8 of the NCERT Class 10 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 8: Introduction to Trigonometry

Introduction

Trigonometry means the measurement of the sides and angles of a triangle. It began with astronomers who needed to find distances they could not measure directly — like the distance from the Earth to the Moon. In this chapter we work only with acute angles in a right-angled triangle and learn six ratios, their values for special angles, and three identities that connect them.

Key Concepts & Theorems

1. The six trigonometric ratios

In right-angled triangle ABC, right-angled at B, consider the acute angle A. The side opposite A is BC, the side adjacent to A is AB, and AC is the hypotenuse.

sin A = side opposite to ∠A / hypotenuse = BC/AC

cos A = side adjacent to ∠A / hypotenuse = AB/AC

tan A = side opposite / side adjacent = BC/AB

cosec A = 1/sin A = AC/BC

sec A = 1/cos A = AC/AB

cot A = 1/tan A = AB/BC

Also, tan A = sin A / cos A and cot A = cos A / sin A.

Important facts from the book:

"sin A" is one symbol — it is not sin × A.

The value of a ratio depends only on the angle, not on the size of the triangle (similar triangles give the same ratio).

Since the hypotenuse is the longest side, sin A and cos A are always less than 1 for an acute angle; sec A and cosec A are always ≥ 1.

2. Ratios of specific angles (0°, 30°, 45°, 60°, 90°)

The book derives these using an isosceles right triangle (for 45°) and an equilateral triangle of side 2a cut in half (for 30° and 60°).

sin: 0°→0, 30°→1/2, 45°→1/√2, 60°→√3/2, 90°→1

cos: 0°→1, 30°→√3/2, 45°→1/√2, 60°→1/2, 90°→0

tan: 0°→0, 30°→1/√3, 45°→1, 60°→√3, 90°→not defined

cosec: 0°→not defined, 30°→2, 45°→√2, 60°→2/√3, 90°→1

sec: 0°→1, 30°→2/√3, 45°→√2, 60°→2, 90°→not defined

cot: 0°→not defined, 30°→√3, 45°→1, 60°→1/√3, 90°→0

Memory tip: write 0, 1, 2, 3, 4 → divide by 4 → take square roots: √(0/4), √(1/4), √(2/4), √(3/4), √(4/4) gives sin 0°, 30°, 45°, 60°, 90°. Reverse the order for cos.

As A increases from 0° to 90°, sin A increases from 0 to 1 and cos A decreases from 1 to 0.

3. Trigonometric identities

An equation that is true for all values of the angle is an identity. Dividing AB² + BC² = AC² by AC², AB² and BC² gives:

sin²A + cos²A = 1 (for 0° ≤ A ≤ 90°)

1 + tan²A = sec²A (for 0° ≤ A < 90°)

1 + cot²A = cosec²A (for 0° < A ≤ 90°)

Using these, any one ratio can be written in terms of any other (Example 9: cos A = √(1 – sin²A), tan A = sin A/√(1 – sin²A), sec A = 1/√(1 – sin²A)).

Methods Step by Step

Finding all ratios when one is given:

Draw a right triangle and mark the angle.

Use the given ratio to label two sides as k and mk (any positive k).

Find the third side with Pythagoras.

Write the remaining ratios.

Proving an identity: start from the more complicated side, convert everything to sin and cos, use sin²A + cos²A = 1, and simplify to the other side.

Worked Examples

Given tan A = 4/3, find the other ratios. Take BC = 4k, AB = 3k. Then AC = 5k. So sin A = 4/5, cos A = 3/5, cosec A = 5/4, sec A = 5/3, cot A = 3/4.

In △OPQ right-angled at P, OP = 7 cm and OQ – PQ = 1 cm. Find sin Q, cos Q. OQ² – PQ² = 49 ⇒ (OQ – PQ)(OQ + PQ) = 49 ⇒ OQ + PQ = 49. So OQ = 25, PQ = 24. sin Q = 7/25, cos Q = 24/25.

In △ABC right-angled at B, AB = 5 cm, ∠ACB = 30°. Find BC and AC. tan 30° = AB/BC ⇒ 1/√3 = 5/BC ⇒ BC = 5√3 cm; sin 30° = AB/AC ⇒ 1/2 = 5/AC ⇒ AC = 10 cm.

In △PQR right-angled at Q, PQ = 3 cm, PR = 6 cm. Find ∠QPR and ∠PRQ. sin R = PQ/PR = 3/6 = 1/2 ⇒ ∠PRQ = 30°, so ∠QPR = 60°.

If sin(A – B) = 1/2 and cos(A + B) = 1/2, find A and B. A – B = 30°, A + B = 60° ⇒ A = 45°, B = 15°.

Prove sec A (1 – sin A)(sec A + tan A) = 1. LHS = (1/cos A)(1 – sin A)(1 + sin A)/cos A = (1 – sin²A)/cos²A = cos²A/cos²A = 1 = RHS.

Board Exam Focus

Standard-angle evaluation questions (e.g. 2 tan²45° + cos²30° – sin²60° = 2 + 3/4 – 3/4 = 2).

"Given one ratio, find the others" — always draw the triangle.

Identity proofs carry 3–4 marks; show every step.

Two-equation problems like sin(A – B), cos(A + B) to find A and B.

Formula Sheet

sin A = P/H, cos A = B/H, tan A = P/B (P = perpendicular/opposite, B = base/adjacent, H = hypotenuse)

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

tan A = sin A/cos A, cot A = cos A/sin A

sin²A + cos²A = 1

sec²A – tan²A = 1

cosec²A – cot²A = 1

Common Mistakes

Treating sin A as "sin × A" or cancelling the A.

Writing sin 60° = 1/2 (that is sin 30°); mixing up the 30° and 60° values.

Saying tan 90° = ∞ — the book says it is not defined.

Writing sin²A as sin A² — sin²A means (sin A)².

In proofs, working on both sides at once instead of transforming one side into the other.

Forgetting that sin A and cos A can never exceed 1 for acute angles.

💡 Key Learning Points

✓Define sin, cos, tan, cosec, sec and cot of an acute angle in a right-angled triangle using opposite, adjacent and hypotenuse.
✓Understand that a trigonometric ratio depends only on the angle, not on the size of the triangle.
✓Recall and derive the ratios of 0°, 30°, 45°, 60° and 90° (with tan 90°, cot 0° etc. not defined).
✓Use sin²A + cos²A = 1, 1 + tan²A = sec²A and 1 + cot²A = cosec²A to prove identities.
✓Express any trigonometric ratio in terms of another and find all ratios when one is given.

👨‍🏫 Teaching Tips

→Have students measure sides of two similar right triangles with the same acute angle and compute sin A to see the ratio stays constant.
→Derive the 30°/60° values live from an equilateral triangle of side 2a split by an altitude — never let students only memorise the table.
→Teach the 0-1-2-3-4 ÷ 4 then square-root trick to rebuild the sin row, and reverse it for cos.
→Use a colour code for opposite, adjacent and hypotenuse, and re-label them each time the reference angle changes.
→For identity proofs, model the 'convert to sin and cos' strategy and insist on starting from one side only.
→Give quick daily oral drills of standard-angle values and simple expressions like 2 sin 30° cos 30°.
→Pair students to create 'given one ratio' problems for each other using Pythagorean triples (3-4-5, 5-12-13, 7-24-25).

📋 Assessment Questions

Given sin A = 3/4, can the student find cos A and tan A correctly?
Can the student evaluate expressions such as 2 tan²45° + cos²30° – sin²60°?
Can the student prove an identity step by step starting from one side?
Can the student explain why sin A < 1 and sec A ≥ 1 for acute A?
Can the student solve for A and B from equations like sin(A – B) = 1/2, cos(A + B) = 1/2?

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