Chapter 8

Class 10 Mathematics - Chapter 8: Introduction to Trigonometry
Chapter 8: Introduction to Trigonometry
1. Trigonometric Ratios
For a right-angled triangle $ABC$ right-angled at $B$ with acute angle $\theta = \angle C$:

$$\sin\theta = \frac{\text{Side opposite to angle }\theta}{\text{Hypotenuse}} = \frac{AB}{AC}$$ $$\cos\theta = \frac{\text{Side adjacent to angle }\theta}{\text{Hypotenuse}} = \frac{BC}{AC}$$ $$\tan\theta = \frac{\text{Side opposite to angle }\theta}{\text{Side adjacent to angle }\theta} = \frac{AB}{BC}$$
Reciprocal Ratios: $$\csc\theta = \frac{1}{\sin\theta} = \frac{AC}{AB}$$ $$\sec\theta = \frac{1}{\cos\cos\theta} = \frac{AC}{BC}$$ $$\cot\theta = \frac{1}{\tan\theta} = \frac{BC}{AB}$$
A B C θ Opposite (AB) Adjacent (BC) Hypotenuse (AC)
Example 1.1: Calculating Trigonometric Ratios

Given $\tan A = \frac{4}{3}$, find the other trigonometric ratios of $\angle A$.

Solution:
Let $\text{Opposite} = 4k$ and $\text{Adjacent} = 3k$.
By Pythagoras theorem: $$\text{Hypotenuse}^2 = (4k)^2 + (3k)^2$$ $$\text{Hypotenuse}^2 = 16k^2 + 9k^2 = 25k^2$$ $$\text{Hypotenuse} = 5k$$ Now evaluate the ratios: $$\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{4k}{5k} = \frac{4}{5}$$ $$\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{3k}{5k} = \frac{3}{5}$$ $$\cot A = \frac{1}{\tan A} = \frac{3}{4}$$ $$\csc A = \frac{1}{\sin A} = \frac{5}{4}$$ $$\sec A = \frac{1}{\cos A} = \frac{5}{3}$$

2. Trigonometric Values of Specific Angles
Standard values of trigonometric ratios for angles $0^\circ, 30^\circ, 45^\circ, 60^\circ,$ and $90^\circ$.
$\angle A$ $0^\circ$ $30^\circ$ $45^\circ$ $60^\circ$ $90^\circ$
$\sin A$ $0$ $\frac{1}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{\sqrt{3}}{2}$ $1$
$\cos A$ $1$ $\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{1}{2}$ $0$
$\tan A$ $0$ $\frac{1}{\sqrt{3}}$ $1$ $\sqrt{3}$ Not Defined
$\csc A$ Not Defined $2$ $\sqrt{2}$ $\frac{2}{\sqrt{3}}$ $1$
$\sec A$ $1$ $\frac{2}{\sqrt{3}}$ $\sqrt{2}$ $2$ Not Defined
$\cot A$ Not Defined $\sqrt{3}$ $1$ $\frac{1}{\sqrt{3}}$ $0$
Example 2.1: Evaluating Expression with Specific Angles

Evaluate: $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$

Solution:
Substitute values from the standard table: $$\sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}$$ $$\sin 30^\circ = \frac{1}{2}, \quad \cos 60^\circ = \frac{1}{2}$$ Substitute into the expression: $$=\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$$ $$= \frac{3}{4} + \frac{1}{4}$$ $$= \frac{4}{4} = 1$$ The value is $1$.

3. Trigonometric Identities
Fundamental Identities: For any angle $A$ ($0^\circ \leq A \leq 90^\circ$):
  1. $$\sin^2 A + \cos^2 A = 1$$
  2. $$1 + \tan^2 A = \sec^2 A$$
  3. $$1 + \cot^2 A = \csc^2 A$$
Quotient Relations: $$\tan A = \frac{\sin A}{\cos A}, \quad \cot A = \frac{\cos A}{\sin A}$$
Example 3.1: Proving Trigonometric Identity

Prove that: $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$

Solution:
Start with the Left Hand Side (LHS): $$\text{LHS} = \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A}$$ Take common denominator: $$= \frac{\cos^2 A + (1 + \sin A)^2}{\cos A (1 + \sin A)}$$ Expand the numerator: $$= \frac{\cos^2 A + 1 + 2\sin A + \sin^2 A}{\cos A (1 + \sin A)}$$ Use $\sin^2 A + \cos^2 A = 1$: $$= \frac{(\sin^2 A + \cos^2 A) + 1 + 2\sin A}{\cos A (1 + \sin A)}$$ $$= \frac{1 + 1 + 2\sin A}{\cos A (1 + \sin A)}$$ $$= \frac{2 + 2\sin A}{\cos A (1 + \sin A)}$$ Factor out 2 in numerator: $$= \frac{2(1 + \sin A)}{\cos A (1 + \sin A)}$$ Cancel $(1 + \sin A)$: $$= \frac{2}{\cos A} = 2 \sec A = \text{RHS}$$ Hence proved.