Class 10 Maths Chapter 8: Introduction to Trigonometry

Introduction to Trigonometry is Chapter 8 of NCERT Class 10 Maths. It introduces the six trigonometric ratios of an acute angle in a right triangle, their values for standard angles, and the basic identities that connect them. Watch the thirteen video lessons below, then use these notes, value table and solved examples to revise.

What you will learn

How to name the opposite side, adjacent side and hypotenuse for a given angle
The six trigonometric ratios: sin, cos, tan, cosec, sec and cot
Finding all ratios when one ratio is given
Values of the ratios for 0°, 30°, 45°, 60° and 90°
The three trigonometric identities and how to prove results using them

Video Lessons

Introduction to Trigonometry - Part 1

Introduction to Trigonometry - Part 2

Introduction to Trigonometry - Part 3

Introduction to Trigonometry - Part 4

Introduction to Trigonometry - Part 5

Introduction to Trigonometry - Part 6

Introduction to Trigonometry - Part 7

Introduction to Trigonometry - Part 8

Introduction to Trigonometry - Part 9

Introduction to Trigonometry - Part 10

Introduction to Trigonometry - Part 11

Introduction to Trigonometry - Exam Practice Part 1

Introduction to Trigonometry - Exam Practice Part 2

Key Concepts and Formulas

Sides of a Right Triangle

The names of the sides depend on which acute angle you are looking at, so always mark the angle first.

Right triangle labelled opposite, adjacent and hypotenuse for angle A with sides 5, 12 and 13 for trigonometric ratios — Sides named for angle A, with sin, cos and tan

Hypotenuse: the side opposite the right angle (the longest side)
Opposite side (perpendicular): the side facing the chosen angle
Adjacent side (base): the side next to the chosen angle, other than the hypotenuse

The Six Ratios

Each ratio compares two sides and depends only on the angle, not on the size of the triangle.

sin A = opposite / hypotenuse
cos A = adjacent / hypotenuse
tan A = opposite / adjacent
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

For an acute angle, sin A and cos A are always less than 1, while sec A and cosec A are greater than 1

Finding Ratios from One Given Ratio

Draw a right triangle, label two sides from the given ratio, find the third side using the Pythagoras relation and read off any other ratio.

If sin A = p/h:
opposite = pk, hypotenuse = hk
adjacent = k√(h² − p²)

Values for Standard Angles

These values come from a 45°-45°-90° triangle and from half of an equilateral triangle. Memorise them well, as they appear in almost every paper.

Angle:  0°   30°    45°    60°    90°
sin:    0    1/2    1/√2   √3/2   1
cos:    1    √3/2   1/√2   1/2    0
tan:    0    1/√3   1      √3     not defined

A 45-45-90 triangle and half an equilateral triangle showing where standard trigonometric values come from — Two triangles behind the standard angle values

cosec, sec and cot are the reciprocals of these values; cosec 0° and cot 0° are not defined, and sec 90° is not defined

Trigonometric Identities

An identity is true for every value of the angle for which both sides are defined. All three identities come from dividing the Pythagoras relation by a suitable side.

sin² A + cos² A = 1       (0° ≤ A ≤ 90°)
1 + tan² A = sec² A       (0° ≤ A < 90°)
1 + cot² A = cosec² A     (0° < A ≤ 90°)

Strategy for Proving Identities

Identity proofs carry 3 to 5 marks in the CBSE board exam, and a steady approach helps more than tricks.

Start from the more complicated side and simplify it towards the other side
Convert everything into sin and cos when you are stuck
Look for a chance to use a² − b² = (a − b)(a + b) or to rationalise by multiplying with a conjugate

Solved Examples

Example 1: If sin A = 5/13, find cos A and tan A.

Opposite = 5, hypotenuse = 13
Adjacent = √(13² − 5²) = √(169 − 25)
Adjacent = √144 = 12
cos A = 12/13, tan A = 5/12
∴ cos A = 12/13, tan A = 5/12

Example 2: If 5 tan θ = 12, find (sin θ + cos θ)/(sin θ − cos θ).

tan θ = 12/5
Divide numerator and denominator by cos θ:
= (tan θ + 1)/(tan θ − 1)
= (12/5 + 1)/(12/5 − 1)
= (17/5)/(7/5)
∴ Value = 17/7

Example 3: In ΔPQR right-angled at Q, PQ = 7 cm and PR = 25 cm. Find sin P, cos P and tan R.

QR = √(PR² − PQ²) = √(625 − 49)
QR = √576 = 24 cm
sin P = QR/PR = 24/25
cos P = PQ/PR = 7/25
tan R = PQ/QR = 7/24
∴ sin P = 24/25, cos P = 7/25, tan R = 7/24

Example 4: Evaluate 4 cos² 60° + 3 sec² 30° − 2 tan² 45°.

cos 60° = 1/2, sec 30° = 2/√3, tan 45° = 1
= 4 × (1/4) + 3 × (4/3) − 2 × 1
= 1 + 4 − 2
∴ Value = 3

Example 5: If sin(A + B) = 1 and cos(A − B) = √3/2, where 0° < A + B ≤ 90° and A > B, find A and B.

sin(A + B) = 1 = sin 90°, so A + B = 90°
cos(A − B) = √3/2 = cos 30°, so A − B = 30°
Adding: 2A = 120°, so A = 60°
B = 90° − 60° = 30°
∴ A = 60°, B = 30°

Example 6: If sec θ + tan θ = 3, find sec θ − tan θ and sec θ.

sec² θ − tan² θ = 1
(sec θ + tan θ)(sec θ − tan θ) = 1
3(sec θ − tan θ) = 1
sec θ − tan θ = 1/3
Adding: 2 sec θ = 3 + 1/3 = 10/3
∴ sec θ − tan θ = 1/3, sec θ = 5/3

Example 7: Prove that sin θ/(1 − cos θ) = (1 + cos θ)/sin θ.

LHS = sin θ/(1 − cos θ)
= sin θ(1 + cos θ)/[(1 − cos θ)(1 + cos θ)]
= sin θ(1 + cos θ)/(1 − cos² θ)
= sin θ(1 + cos θ)/sin² θ
= (1 + cos θ)/sin θ
∴ LHS = RHS

Common Mistakes to Avoid

Writing sin A as "sin × A"; sin A is a single value, so you cannot cancel "sin" from sin A/sin B.
Confusing sin² A with sin A²; sin² A means (sin A)².
Mixing up the values of sin 30° and sin 60°; remember that sin increases with the angle, so sin 30° = 1/2 is the smaller one.
Working on both sides of an identity at the same time without care; show the steps on one side and reach the other.

Quick Revision

sin = opp/hyp, cos = adj/hyp, tan = opp/adj
cosec = 1/sin, sec = 1/cos, cot = 1/tan
sin 30° = cos 60° = 1/2
sin 45° = cos 45° = 1/√2, tan 45° = 1
sin² A + cos² A = 1
1 + tan² A = sec² A
1 + cot² A = cosec² A