support@tutormate.in   022 6236 4602

# Introduction to trigonometry

Tutormate > CBSE Syllabus-Class 10th Maths > Introduction to trigonometry

## 08 INTRODUCTION TO TRIGONOMETRY

Tigonometric Ratios :

In right angled triangle $\text{ABC}$

Relation between ratios:

Trigonometric ratios of Specific angles:

 $\text{θ}$ $\text{0°}$ $\text{30°}$ $\text{45°}$ $\text{60°}$ $\text{90°}$ 0 $\frac{1}{2}$ $\frac{1}{\sqrt{\text{2}}}$ $\frac{\sqrt{\text{3}}}{\text{2}}$ 1 1 $\frac{\sqrt{\text{3}}}{\text{2}}$ $\frac{1}{\sqrt{\text{2}}}$ $\frac{\text{1}}{\text{2}}$ 0 0 $\frac{\text{1}}{\sqrt{\text{3}}}$ 1 $\sqrt{\text{3}}$ Not Defined Not defined 2 $\sqrt{\text{2}}$ $\frac{\text{2}}{\sqrt{\text{3}}}$ 1 1 $\frac{\text{2}}{\sqrt{\text{3}}}$ $\sqrt{\text{2}}$ 2 Not defined Not defined $\sqrt{\text{3}}$ 1 $\frac{1}{\sqrt{\text{3}}}$

Trigonometric Ratios of Complimentary Angles:

Trigonometric Identities:

$1+{\mathrm{tan}}^{2}\theta ={\mathrm{sec}}^{2}\theta$

$1+{\mathrm{cot}}^{2}\theta ={\mathrm{csc}}^{2}\theta$

Proofs: In a $\Delta \text{ABC}$,

1) By Pythagoras theorem

Dividing by $A{C}^{2}$, we get

2) By Pythagoras theorem

Dividing by ${\text{BC}}^{\text{2}}$, we get

3) By Pythagoras theorem

Dividing by ${\text{AB}}^{\text{2}}$, we get

Different forms of identities:

1) or or

2) $1+{\mathrm{tan}}^{2}\theta ={\mathrm{sec}}^{2}\theta$ or ${\mathrm{sec}}^{2}\theta -{\mathrm{tan}}^{2}\theta =1$ or ${\mathrm{tan}}^{2}\theta ={\mathrm{sec}}^{2}\theta -1$

3) $1+{\mathrm{cot}}^{2}\theta =co{\mathrm{sec}}^{2}\theta$ or $co{\mathrm{sec}}^{2}\theta -{\mathrm{cot}}^{2}\theta =1$ or ${\mathrm{cot}}^{2}\theta =co{\mathrm{sec}}^{2}\theta -1$

.