CBSE Class 10 Mathematics Trigonometry Notes Set 14

Introduction to Trigonometry and Trigonometric Ratios

Trigonometry means the science which deals with the measurement of triangles. The word 'trigonometry' is derived from the Greek words 'trigon' and 'metron' and it means 'measuring the sides and angles of a triangle'.

Trigonometric Ratios

A right angled triangle is shown in the figure. \( \angle B \) is of \( 90^\circ \). The side opposite to \( \angle B \) is called the hypotenuse. There are two other angles, \( \angle A \) and \( \angle C \). If we consider \( \angle C \) as \( \theta \), then the side opposite to this angle is called the perpendicular and the side adjacent to \( \theta \) is called the base.

Six Trigonometric Ratios

\[ \sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{P}{H} = \frac{AB}{AC}, \qquad \operatorname{cosec}\theta = \frac{H}{P} = \frac{AC}{AB} \]

\[ \cos\theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{B}{H} = \frac{BC}{AC}, \qquad \sec\theta = \frac{H}{B} = \frac{AC}{BC} \]

\[ \tan\theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{P}{B} = \frac{AB}{BC}, \qquad \cot\theta = \frac{B}{P} = \frac{BC}{AB} \]

Interrelationship in Basic Trigonometric Ratios

\[ \tan\theta = \frac{1}{\cot\theta}, \quad \cos\theta = \frac{1}{\sec\theta}, \quad \sin\theta = \frac{1}{\operatorname{cosec}\theta} \]

\[ \cot\theta = \frac{1}{\tan\theta}, \quad \sec\theta = \frac{1}{\cos\theta}, \quad \operatorname{cosec}\theta = \frac{1}{\sin\theta} \]

We also observe that \( \tan\theta = \dfrac{\sin\theta}{\cos\theta} \) and \( \cot\theta = \dfrac{\cos\theta}{\sin\theta} \).

Solved Examples

Example 2. If \( \tan\theta = \dfrac{m}{n} \), then find \( \sin\theta \).
Answer:
\( \tan\theta = \dfrac{P}{B} = \dfrac{m}{n} \). Let P = mα and B = nα (where α is some constant).
\( H^2 = P^2 + B^2 = \alpha^2(m^2 + n^2) \Rightarrow H = \alpha\sqrt{m^2 + n^2} \).
\( \sin\theta = \dfrac{P}{H} = \dfrac{m\alpha}{\alpha\sqrt{m^2 + n^2}} = \dfrac{m}{\sqrt{m^2 + n^2}} \).

Check Your Progress

Check Your Progress 1. If \( \sin\theta = \dfrac{3}{5} \), find all the other ratios.
Answer:
P = 3, H = 5, so B = 4. \( \cos\theta = \dfrac{4}{5},\ \tan\theta = \dfrac{3}{4},\ \cot\theta = \dfrac{4}{3},\ \sec\theta = \dfrac{5}{4},\ \operatorname{cosec}\theta = \dfrac{5}{3} \).

Check Your Progress 2. If \( \tan\theta = \dfrac{12}{5} \), find the value of \( \dfrac{\sin\theta + \cot\theta}{\cos\theta + \operatorname{cosec}\theta} \).
Answer:
P = 12, B = 5, H = 13. \( \sin\theta = \tfrac{12}{13} \), \( \cot\theta = \tfrac{5}{12} \), \( \cos\theta = \tfrac{5}{13} \), \( \operatorname{cosec}\theta = \tfrac{13}{12} \). The value is \( \dfrac{\frac{144 + 65}{156}}{\frac{60 + 169}{156}} = \dfrac{209}{229} \).

Check Your Progress 3. If \( \operatorname{cosec}A = \dfrac{5}{3} \), then prove that \( \tan^2A - \sin^2A = \sin^4A\sec^2A \).
Answer:
P = 3, H = 5, B = 4. \( \tan A = \tfrac{3}{4} \), \( \sin A = \tfrac{3}{5} \), \( \sec A = \tfrac{5}{4} \). LHS \( = \tfrac{9}{16} - \tfrac{9}{25} = \tfrac{81}{400} \). RHS \( = \tfrac{81}{625} \times \tfrac{25}{16} = \tfrac{81}{400} \). Hence proved.

Check Your Progress 4. In \( \triangle ABC \), right-angled at B, AC + AB = 9 cm and BC = 3 cm. Determine the value of cot C and cosec C.
Answer:
Let AB = x, so AC = 9 − x. \( (9 - x)^2 = x^2 + 9 \Rightarrow 81 - 18x = 9 \Rightarrow x = 4 \), AC = 5. For angle C: perpendicular AB = 4, base BC = 3. So \( \cot C = \dfrac{3}{4} \) and \( \operatorname{cosec}C = \dfrac{5}{4} \).

Check Your Progress 5. Find the value of \( \sin\theta \cdot \sec\theta \cdot \operatorname{cosec}\theta \cdot \cos\theta \).
Answer:
\( \sin\theta \cdot \dfrac{1}{\cos\theta} \cdot \dfrac{1}{\sin\theta} \cdot \cos\theta = \) 1.