ICSE Class 10 Maths Chapter 25 Trigonometrical Identities

Unit 6 Trigonometry

Chapter 25 Trigonometrical Identities

Points To Remember

1. Trigonometrical Ratios or T-Ratios

Let A be an acute angle of a right triangle ABC in which angle B = 90 degrees. Then, Base = AB, Perpendicular = BC and Hypotenuse = AC.

For acute angle A = θ, we define

(i) sine θ = Perpendicular / Hypotenuse = \(\sin \theta = \frac{BC}{AC}\)

(ii) cosine θ = Base / Hypotenuse = \(\cos \theta = \frac{AB}{AC}\)

(iii) tangent θ = Perpendicular / Base = \(\tan \theta = \frac{BC}{AB}\)

(iv) cotangent θ = Base / Perpendicular = \(\cot \theta = \frac{AB}{BC}\)

(v) secant θ = Hypotenuse / Base = \(\sec \theta = \frac{AC}{AB}\)

(vi) cosecant θ = Hypotenuse / Perpendicular = \(\csc \theta = \frac{AC}{BC}\)

2. (A) Relations Between T-Ratios (Theorems)

Theorem 1. For an acute angle A, prove that:

(i) \(\csc A = \frac{1}{\sin A}\)

(ii) \(\sec A = \frac{1}{\cos A}\)

(iii) \(\cot A = \frac{1}{\tan A}\)

Proof

We have,

(i) \(\csc A = \frac{AC}{BC}\) and \(\sin A = \frac{BC}{AC}\)

\(\therefore \csc A = \frac{1}{\sin A}\)

(ii) \(\sec A = \frac{AC}{AB}\) and \(\cos A = \frac{AB}{AC}\)

\(\therefore \sec A = \frac{1}{\cos A}\)

(iii) \(\cot A = \frac{AB}{BC}\) and \(\tan A = \frac{BC}{AB}\)

\(\therefore \cot A = \frac{1}{\tan A}\)

(B) Quotient Relations

Theorem 2. For an acute angle A, prove that:

(i) \(\frac{\sin A}{\cos A} = \tan A\)

(ii) \(\frac{\cos A}{\sin A} = \cot A\)

(iii) \(\tan A \cot A = 1\)

Proof

We have,

\(\sin A = \frac{BC}{AC}\) and \(\cos A = \frac{AB}{AC}\)

(i) \(\frac{\sin A}{\cos A} = \left(\frac{BC}{AC} \cdot \frac{AC}{AB}\right) = \frac{BC}{AB} = \tan A\)

(ii) \(\frac{\cos A}{\sin A} = \left(\frac{AB}{AC} \cdot \frac{AC}{BC}\right) = \frac{AB}{BC} = \cot A\)

(iii) \(\tan A \cot A = \left(\frac{BC}{AB} \cdot \frac{AB}{BC}\right) = 1\)

(C) Square Relations

Theorem 3. For an acute angle A, prove that:

(i) \(\sin^2 A + \cos^2 A = 1\)

(ii) \(1 + \tan^2 A = \sec^2 A\)

(iii) \(1 + \cot^2 A = \csc^2 A\)

Proof

We have,

(i) \(\sin^2 A + \cos^2 A = \left(\frac{BC}{AC}\right)^2 + \left(\frac{AB}{AC}\right)^2\)

\(= \frac{BC^2}{AC^2} + \frac{AB^2}{AC^2}\)

\(= \frac{BC^2 + AB^2}{AC^2} = \frac{AC^2}{AC^2} = 1\)

[Since \(BC^2 + AB^2 = AC^2\)]

Hence, \(\sin^2 A + \cos^2 A = 1\)

(ii) \(1 + \tan^2 A = 1 + \left(\frac{BC}{AB}\right)^2 = 1 + \frac{BC^2}{AB^2}\)

\(= \frac{AB^2 + BC^2}{AB^2} = \frac{AC^2}{AB^2}\)

[Since \(AB^2 + BC^2 = AC^2\)]

(iii) \(1 + \cot^2 A = 1 + \left(\frac{AB}{BC}\right)^2 = 1 + \frac{AB^2}{BC^2}\)

\(= \frac{BC^2 + AB^2}{BC^2} = \frac{AC^2}{BC^2} = \left(\frac{AC}{BC}\right)^2 = \csc^2 A\)

Hence, \(1 + \cot^2 A = \csc^2 A\)

3. Trigonometrical Ratios of Complementary Angles

Complementary Angles

Two angles are said to be complementary, if the sum of their measures is 90 degrees. Thus, A and (90 degrees - A) are complementary angles.

4. T-Ratios of Complementary Angles

Consider a right triangle ABC in which angle BAC = 90 degrees.

Therefore, angle ACB = (90 degrees - A).

Thus, we have:

(i) \(\sin(90° - A) = \frac{AB}{AC} = \cos A\)

(ii) \(\cos(90° - A) = \frac{BC}{AC} = \sin A\)

(iii) \(\tan(90° - A) = \frac{AB}{BC} = \cot A\)

(iv) \(\csc(90° - A) = \frac{1}{\sin(90° - A)} = \frac{\cos A}{1} = \sec A\)

(v) \(\sec(90° - A) = \frac{1}{\cos(90° - A)} = \frac{\sin A}{1} = \csc A\)

(vi) \(\cot(90° - A) = \frac{1}{\tan(90° - A)} = \frac{\cot A}{1} = \tan A\)

Using Trigonometric Tables

Table Showing T-Ratios of 0 degrees, 30 degrees, 45 degrees, 60 degrees and 90 degrees

Teacher's Note

Trigonometric ratios form the foundation for understanding angles and distances, which we use every day in activities like navigation, construction, and even sports analytics.

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