Introduction to Trigonometry - Class 10 Maths

1. Definition of Trigonometry

The word trigonometry is derived from the Greek words 'tri' (meaning three), 'gon' (meaning sides) and 'metron' (meaning measure). Trigonometry is the branch of mathematics that studies relationships between side lengths and angles of triangles.

2. Mathematical Definition of Trigonometric Ratios

Consider a right-angled triangle $ABC$, right-angled at $B$. For acute angle $A$ ($\angle CAB = \theta$):

Side opposite to angle $A$ $= BC$ (Perpendicular, $p$)
Side adjacent to angle $A$ $= AB$ (Base, $b$)
Hypotenuse $= AC$ (Side opposite to $90^\circ$, $h$)

Right $\triangle ABC$ showing sides with respect to acute angle $\angle A = \theta$.

The 6 Trigonometric Ratios $$\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{p}{h} \qquad\qquad \csc\theta = \frac{\text{Hypotenuse}}{\text{Opposite}} = \frac{h}{p}$$ $$\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{b}{h} \qquad\qquad \sec\theta = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{h}{b}$$ $$\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{p}{b} \qquad\qquad \cot\theta = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{b}{p}$$

3. Reciprocal Relations

By comparing the ratios directly from their geometric definitions, each ratio pairs with its exact multiplicative inverse:

Reciprocal Formulas $$\csc\theta = \frac{1}{\sin\theta} \iff \sin\theta \cdot \csc\theta = 1$$ $$\sec\theta = \frac{1}{\cos\theta} \iff \cos\theta \cdot \sec\theta = 1$$ $$\cot\theta = \frac{1}{\tan\theta} \iff \tan\theta \cdot \cot\theta = 1$$

4. $\tan$ and $\cot$ in Terms of $\sin$, $\cos$, $\csc$, $\sec$

Expressing $\tan\theta$ and $\cot\theta$ as quotients of primary ratios:

Quotient Relations $$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{1}{\csc\theta}}{\frac{1}{\sec\theta}} = \frac{\sec\theta}{\csc\theta}$$ $$\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{\frac{1}{\sec\theta}}{\frac{1}{\csc\theta}} = \frac{\csc\theta}{\sec\theta}$$

5. Trigonometric Identities and Their Proofs

An equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all values of the angles for which the ratios are defined.

Right $\triangle ABC$, right-angled at $B$. By Pythagoras theorem: $AB^2 + BC^2 = AC^2$.

Proof of Identity 1: $\sin^2 A + \cos^2 A = 1$

In right $\triangle ABC$, by Pythagoras Theorem: $$AB^2 + BC^2 = AC^2 \quad \text{--- (1)}$$ Dividing every term of (1) by $AC^2$: $$\frac{AB^2}{AC^2} + \frac{BC^2}{AC^2} = \frac{AC^2}{AC^2} \implies \left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = 1$$ Since $\cos A = \frac{AB}{AC}$ and $\sin A = \frac{BC}{AC}$: $$\cos^2 A + \sin^2 A = 1 \implies \mathbf{\sin^2 A + \cos^2 A = 1}$$ This holds for all $0^\circ \le A \le 90^\circ$.

Proof of Identity 2: $1 + \tan^2 A = \sec^2 A$

Dividing every term of (1) by $AB^2$: $$\frac{AB^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AC^2}{AB^2} \implies 1 + \left(\frac{BC}{AB}\right)^2 = \left(\frac{AC}{AB}\right)^2$$ Since $\tan A = \frac{BC}{AB}$ and $\sec A = \frac{AC}{AB}$: $$\mathbf{1 + \tan^2 A = \sec^2 A} \quad (\text{or } \sec^2 A - \tan^2 A = 1)$$ Defined for $0^\circ \le A < 90^\circ$ (since $\tan 90^\circ$ is not defined).

Proof of Identity 3: $1 + \cot^2 A = \csc^2 A$

Dividing every term of (1) by $BC^2$: $$\frac{AB^2}{BC^2} + \frac{BC^2}{BC^2} = \frac{AC^2}{BC^2} \implies \left(\frac{AB}{BC}\right)^2 + 1 = \left(\frac{AC}{BC}\right)^2$$ Since $\cot A = \frac{AB}{BC}$ and $\csc A = \frac{AC}{BC}$: $$\mathbf{1 + \cot^2 A = \csc^2 A} \quad (\text{or } \csc^2 A - \cot^2 A = 1)$$ Defined for $0^\circ < A \le 90^\circ$ (since $\cot 0^\circ$ is not defined).

6. Values for 45° with Geometric Proof

Right-angled isosceles triangle $ABC$ with $\angle A = \angle C = 45^\circ$.

Proof

1. Consider an isosceles right $\triangle ABC$ right-angled at $B$, where $\angle A = 45^\circ$.
2. Since $\angle A + \angle B + \angle C = 180^\circ$, $\angle C = 180^\circ - 90^\circ - 45^\circ = 45^\circ$.
3. Since $\angle A = \angle C$, sides opposite to equal angles are equal: $BC = AB = a$.
4. By Pythagoras Theorem: $$AC = \sqrt{AB^2 + BC^2} = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}$$ 5. Evaluating all ratios for $45^\circ$: $$\sin 45^\circ = \frac{BC}{AC} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}} \qquad\qquad \cos 45^\circ = \frac{AB}{AC} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}}$$ $$\tan 45^\circ = \frac{BC}{AB} = \frac{a}{a} = 1 \qquad\qquad\qquad \cot 45^\circ = \frac{1}{\tan 45^\circ} = 1$$ $$\csc 45^\circ = \frac{1}{\sin 45^\circ} = \sqrt{2} \qquad\qquad\qquad \sec 45^\circ = \frac{1}{\cos 45^\circ} = \sqrt{2}$$

7. Values for 30° and 60° with Geometric Proof

Equilateral $\triangle ABC$ of side $2a$ with perpendicular altitude $AD$.

Proof

1. Consider an equilateral triangle $ABC$ with each side of length $2a$, so $\angle A = \angle B = \angle C = 60^\circ$.
2. Draw $AD \perp BC$. In $\triangle ABD$ and $\triangle ACD$: $$\angle ADB = \angle ADC = 90^\circ, \quad AB = AC = 2a, \quad AD = AD \text{ (Common)}$$ Hence $\triangle ABD \cong \triangle ACD$ (RHS rule).
3. This gives $BD = CD = a$, and $\angle BAD = \angle CAD = 30^\circ$.
4. In right $\triangle ABD$, by Pythagoras Theorem: $$AD = \sqrt{AB^2 - BD^2} = \sqrt{(2a)^2 - a^2} = \sqrt{4a^2 - a^2} = \sqrt{3a^2} = a\sqrt{3}$$
Values for 30°: (In $\triangle ABD$, with angle $\angle BAD = 30^\circ$, opposite $= BD = a$, adjacent $= AD = a\sqrt{3}$, hypotenuse $= AB = 2a$): $$\sin 30^\circ = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2} \qquad\qquad \cos 30^\circ = \frac{AD}{AB} = \frac{a\sqrt{3}}{2a} = \frac{\sqrt{3}}{2}$$ $$\tan 30^\circ = \frac{BD}{AD} = \frac{a}{a\sqrt{3}} = \frac{1}{\sqrt{3}} \qquad\quad \csc 30^\circ = 2, \quad \sec 30^\circ = \frac{2}{\sqrt{3}}, \quad \cot 30^\circ = \sqrt{3}$$
Values for 60°: (In $\triangle ABD$, with angle $\angle B = 60^\circ$, opposite $= AD = a\sqrt{3}$, adjacent $= BD = a$, hypotenuse $= AB = 2a$): $$\sin 60^\circ = \frac{AD}{AB} = \frac{a\sqrt{3}}{2a} = \frac{\sqrt{3}}{2} \qquad\qquad \cos 60^\circ = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2}$$ $$\tan 60^\circ = \frac{AD}{BD} = \frac{a\sqrt{3}}{a} = \sqrt{3} \qquad\quad \csc 60^\circ = \frac{2}{\sqrt{3}}, \quad \sec 60^\circ = 2, \quad \cot 60^\circ = \frac{1}{\sqrt{3}}$$

8. Geometric Explanation for 0° and 90°

Limiting behavior of sides as acute angle $\angle A$ approaches $0^\circ$ and $90^\circ$.

Explanation for 0°

As $\angle A$ gets smaller and smaller until it approaches $0^\circ$:
1. The length of perpendicular side $BC$ decreases and approaches $0$ ($BC \to 0$).
2. The length of hypotenuse $AC$ decreases and coincides with base $AB$ ($AC \to AB$).
Substituting these limits into our ratio definitions: $$\sin 0^\circ = \frac{BC}{AC} \to \frac{0}{AB} = \mathbf{0} \qquad\qquad \cos 0^\circ = \frac{AB}{AC} \to \frac{AB}{AB} = \mathbf{1}$$ $$\tan 0^\circ = \frac{\sin 0^\circ}{\cos 0^\circ} = \frac{0}{1} = \mathbf{0} \qquad\qquad \sec 0^\circ = \frac{1}{\cos 0^\circ} = \frac{1}{1} = \mathbf{1}$$ $$\csc 0^\circ = \frac{1}{\sin 0^\circ} = \frac{1}{0} \implies \textbf{Not defined} \qquad \cot 0^\circ = \frac{1}{\tan 0^\circ} = \frac{1}{0} \implies \textbf{Not defined}$$

Explanation for 90°

As $\angle A$ increases toward $90^\circ$, point $A$ moves closer and closer to point $B$:
1. The length of adjacent side $AB$ approaches $0$ ($AB \to 0$).
2. The length of hypotenuse $AC$ coincides almost exactly with perpendicular $BC$ ($AC \to BC$).
Substituting these limits: $$\sin 90^\circ = \frac{BC}{AC} \to \frac{BC}{BC} = \mathbf{1} \qquad\qquad \cos 90^\circ = \frac{AB}{AC} \to \frac{0}{BC} = \mathbf{0}$$ $$\tan 90^\circ = \frac{\sin 90^\circ}{\cos 90^\circ} = \frac{1}{0} \implies \textbf{Not defined} \qquad \csc 90^\circ = \frac{1}{\sin 90^\circ} = \frac{1}{1} = \mathbf{1}$$ $$\sec 90^\circ = \frac{1}{\cos 90^\circ} = \frac{1}{0} \implies \textbf{Not defined} \qquad \cot 90^\circ = \frac{\cos 90^\circ}{\sin 90^\circ} = \frac{0}{1} = \mathbf{0}$$

9. Ratio and Value Table Construction Trick

You can construct the complete table of standard angles without memorizing by following these 7 progressive steps:

Step 0 Click "Next Step" to start constructing the table from scratch.

Ratio / Angle	0°	30°	45°	60°	90°
sin A	-	-	-	-	-
cos A	-	-	-	-	-
tan A	-	-	-	-	-
csc A	-	-	-	-	-
sec A	-	-	-	-	-
cot A	-	-	-	-	-

10. Trigonometric Ratios of Complementary Angles

Two angles are said to be complementary if their sum equals $90^\circ$. In a right $\triangle ABC$ right-angled at $B$, $\angle A + \angle C = 90^\circ \implies \angle C = 90^\circ - A$.

Complementary Angle Rules $$\sin(90^\circ - A) = \cos A \qquad\qquad \cos(90^\circ - A) = \sin A$$ $$\tan(90^\circ - A) = \cot A \qquad\qquad \cot(90^\circ - A) = \tan A$$ $$\sec(90^\circ - A) = \csc A \qquad\qquad \csc(90^\circ - A) = \sec A$$

Proof

In right $\triangle ABC$ with $\angle B = 90^\circ$:
With respect to angle $A$: $\sin A = \frac{BC}{AC}$, $\cos A = \frac{AB}{AC}$, $\tan A = \frac{BC}{AB}$.
With respect to angle $C = (90^\circ - A)$: $$\sin(90^\circ - A) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} = \cos A$$ $$\cos(90^\circ - A) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} = \sin A$$ $$\tan(90^\circ - A) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} = \cot A$$ Similarly, reciprocating these yields the rules for $\cot$, $\sec$, and $\csc$.