CBSE Class 10 Mathematics Trigonometry Notes Set 10

Introduction

The word 'Trigonometry' is derived from the Greek words 'tri' meaning three, 'gon' meaning side and 'metron' meaning measure. Hence, trigonometry means the science of measuring triangles. Trigonometry is the study of the relationship between the sides and the angles of a triangle.

Trigonometry has its application in astronomy, geography, surveying, engineering and navigation. In the past, astronomers used it to find the distance of stars and planets from the earth. Even now, the advanced technology used in engineering is based on trigonometric concepts.

In this chapter, we define the trigonometric ratios of angles in terms of the ratios of sides of a right triangle. We also define the trigonometric ratios of the angles \( 0^\circ, 30^\circ, 45^\circ, 60^\circ \) and \( 90^\circ \), and establish some identities involving these ratios.

Right Angle Triangle

(i) A triangle in which one angle is equal to \( 90^\circ \) is known as a right angled triangle.
(ii) The side opposite to \( 90^\circ \) is called the hypotenuse. It is the longest side of the triangle.
(iii) The side opposite to the given acute angle is the perpendicular.
(iv) The remaining third side, that is, the side forming the given angle with the hypotenuse, is called the base.

Here \( \theta \) is the angle under consideration. In \( \triangle ABC \) with \( \angle B = 90^\circ \), the sides for the two acute angles are:

Trigonometrical Ratios (T-Ratios) of an Angle

In \( \triangle ABC \), let \( \angle B = 90^\circ \) and let \( \angle A \) be acute. For \( \angle A \) we have Base \( = AB \), Perpendicular \( = BC \) and Hypotenuse \( = AC \). The T-ratios for \( \angle A \) are defined as:

(i) \( \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AC} \)
(ii) \( \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{AB}{AC} \)
(iii) \( \tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB} \)
(iv) \( \operatorname{cosec} A = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{AC}{BC} \)
(v) \( \sec A = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{AC}{AB} \)
(vi) \( \cot A = \frac{\text{Base}}{\text{Perpendicular}} = \frac{AB}{BC} \)

Thus, there are six trigonometrical ratios based on the three sides of a right angled triangle.

Knowledge Hub

1. \( \sin \theta \) is the sine of the angle \( \theta \).
2. \( \sin \theta \) is NOT \( \sin \times \theta \).
3. Mnemonic: Pandit Badri Prasad / Har Har Bhole / Sona Chandi Tole, that is, \( \sin \theta = \frac{\text{Pandit}}{\text{Har}},\ \cos \theta = \frac{\text{Badri}}{\text{Har}} \) and \( \tan \theta = \frac{\text{Prasad}}{\text{Bhole}} \) (Perpendicular, Base, Hypotenuse).

Relations Between Trigonometric Ratios

Reciprocal relations
(i) \( \operatorname{cosec} \theta = \frac{1}{\sin \theta} \)
(ii) \( \sec \theta = \frac{1}{\cos \theta} \)
(iii) \( \cot \theta = \frac{1}{\tan \theta} \)

Quotient relations. Consider a right angled triangle in which, for an acute angle \( \theta \), \( \sin \theta = \frac{P}{H} \) and \( \cos \theta = \frac{B}{H} \). Now,

\( \frac{\sin \theta}{\cos \theta} = \frac{P}{H} \times \frac{H}{B} = \frac{P}{B} = \tan \theta \) and \( \frac{\cos \theta}{\sin \theta} = \frac{B}{H} \times \frac{H}{P} = \frac{B}{P} = \cot \theta \).

Thus, \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \cot \theta = \frac{\cos \theta}{\sin \theta} \).

Knowledge Hub

1. \( (\sin \theta)^2 \) is written as \( \sin^2 \theta \), \( (\cos \theta)^2 \) as \( \cos^2 \theta \), \( (\sin \theta)^3 \) as \( \sin^3 \theta \), and so on.
2. \( \sin^2 \theta \neq \sin \theta^2 \) and \( \sin 2A \neq 2 \sin A \). Also \( \sin A \neq \sin \times A \), \( \cos A \neq \cos \times A \) and \( \tan A \neq \tan \times A \).
3. Since the hypotenuse is the longest side of a right triangle, the value of \( \sin A \) or \( \cos A \) is always less than 1 (or, in particular, equal to 1).

Solved Examples

Example 1. In a right triangle ABC, right angled at B, the ratio of AB to AC is \( 1 : \sqrt{2} \). Find the value of \( \frac{2 \tan A}{1 - \tan^2 A} \).

Answer:
We have \( AB : AC = 1 : \sqrt{2} \), i.e. \( \frac{AB}{AC} = \frac{1}{\sqrt{2}} \). Let \( AB = x \) and \( AC = \sqrt{2}\,x \).
By Pythagoras theorem, \( AC^2 = AB^2 + BC^2 \Rightarrow (\sqrt{2}\,x)^2 = x^2 + BC^2 \Rightarrow BC^2 = 2x^2 - x^2 = x^2 \Rightarrow BC = x \).
\( \therefore \tan A = \frac{BC}{AB} = \frac{x}{x} = 1 \).
Now, \( \frac{2 \tan A}{1 - \tan^2 A} = \frac{2 \times 1}{1 - 1} = \frac{2}{0} \), which is undefined.

Example 2. If \( \sin A = \frac{3}{5} \), find \( \cos A \) and \( \tan A \).

Answer:
Since \( \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{3}{5} \), Perpendicular \( = BC = 3 \) units and Hypotenuse \( = AC = 5 \) units.
By Pythagoras theorem, \( AC^2 = AB^2 + BC^2 \Rightarrow 5^2 = AB^2 + 3^2 \Rightarrow AB^2 = 25 - 9 = 16 \Rightarrow AB = 4 \) (Base).
\( \therefore \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{4}{5} \) and \( \tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{3}{4} \).

Example 3. If \( \frac{\operatorname{cosec} \theta + \cot \theta}{\operatorname{cosec} \theta - \cot \theta} = \frac{81}{49} \), find \( \frac{\cos \theta + \sin \theta}{\sin \theta - \cos \theta} \).

Answer:
\( \frac{\operatorname{cosec} \theta + \cot \theta}{\operatorname{cosec} \theta - \cot \theta} = \frac{81}{49} \Rightarrow 49 \operatorname{cosec} \theta + 49 \cot \theta = 81 \operatorname{cosec} \theta - 81 \cot \theta \Rightarrow 130 \cot \theta = 32 \operatorname{cosec} \theta \).
\( \Rightarrow 130 \cdot \frac{\cos \theta}{\sin \theta} = 32 \cdot \frac{1}{\sin \theta} \Rightarrow 65 \cos \theta = 16 \Rightarrow \cos \theta = \frac{16}{65} = \frac{\text{Base}}{\text{Hypotenuse}} \).
Applying Pythagoras theorem in \( \triangle ABC \): \( AB^2 = AC^2 - BC^2 = 65^2 - 16^2 = (65 + 16)(65 - 16) = 81 \times 49 \), so \( AB = 9 \times 7 = 63 \). Hence \( \sin \theta = \frac{AB}{AC} = \frac{63}{65} \).
\( \therefore \frac{\cos \theta + \sin \theta}{\sin \theta - \cos \theta} = \frac{\frac{16}{65} + \frac{63}{65}}{\frac{63}{65} - \frac{16}{65}} = \frac{79}{65} \times \frac{65}{47} = \frac{79}{47} \).

Example 4. If \( \tan \theta = \frac{2x(x + 1)}{2x + 1} \), find \( \sin \theta \) and \( \cos \theta \).

Answer:
\( \tan \theta = \frac{2x(x + 1)}{2x + 1} = \frac{P}{B} \). Let \( P = PM = 2x(x + 1) \), \( B = OM = 2x + 1 \) and \( H = OP \).
In \( \triangle PMO \): \( OP^2 = PM^2 + OM^2 = [2x(x + 1)]^2 + (2x + 1)^2 = 4x^4 + 8x^3 + 8x^2 + 4x + 1 = (2x^2 + 2x + 1)^2 \).
\( \Rightarrow OP = 2x^2 + 2x + 1 \).
\( \sin \theta = \frac{PM}{OP} = \frac{2x(x + 1)}{2x^2 + 2x + 1} \) and \( \cos \theta = \frac{OM}{OP} = \frac{2x + 1}{2x^2 + 2x + 1} \).