# Trigonometry Formulas for Class 10 Maths

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Here we are providing trigonometry formulas for class 10 maths. Students are suggested to remember these formulas so that there won’t be any problem in solving questions related to trigonometry.

## Trigonometry Formulas for Class 10 Maths

Trigonometric Ratios (T-RATIOS) of an Acute Angle of a Right Triangle

(i) $${sine}\, \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{y}{r}$$ Note: It is written as $\sin \theta\$.
(ii) $${cosine}\, \theta=\frac{\text { base }}{\text { hypotenuse }}=\frac{x}{r}$$ Note: It is written as $\cos \theta\$.
(iii) $$tangent\, \theta=\frac{\text { perpendicular }}{\text { base }}=\frac{y}{x}$$ Note: it is written as $\tan \theta\$.
(iv) $${cosecant}\, \theta=\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{r}{y}$$ Note: It is written as ${cosec} \theta$.
(v) $${secant} \theta=\frac{\text { hypotenuse }}{\text { base }}=\frac{r}{x}$$ Note: It is written as $\sec \theta\$.
(vi) $$cotangent\, \theta=\frac{\text { base }}{\text { perpendicular }}=\frac{x}{y}$$ Note: It is written as $\cot \theta\$.

RECIPROCAL RELATION

$$(i) \,cosec \theta=\frac{1}{\sin \theta}$$
$$(ii) \,\sec \theta=\frac{1}{\cos \theta}$$
$$(iii)\, \cot \theta=\frac{1}{\tan \theta}$$

POWER OF T-RATIOS

$$(\sin \theta)^2=\sin ^2 \theta ; \quad(\sin \theta)^3=\sin ^3 \theta ; \quad(\cos \theta)^3=\cos ^3 \theta$$

QUOTIENT RELATION OF T-RATIOS

$$(i) \,\tan \theta=\frac{\sin \theta}{\cos \theta}$$
$$(ii) \, \cot \theta=\frac{\cos \theta}{\sin \theta}$$
$$(iii)\, \tan \theta \cdot \cot \theta=1$$

Square Relation or Trigonometric Identities

$$(i) \,\sin ^2 \theta+\cos ^2 \theta=1$$
$$(ii) \,1+\tan ^2 \theta=\sec ^2 \theta$$
$$(iii) \,1+\cot ^2 \theta={cosec} ^2 \theta$$

TABLE FOR T-RATIOS

TRIGONOMETRIC RATIOS OF COMPLEMENTARY ANGLES

$$(i)\,\sin \left(90^{\circ}-\theta\right)=\cos \theta$$
$$(ii)\,\cos \left(90^{\circ}-\theta\right)=\sin \theta$$
$$(iii)\,\tan \left(90^{\circ}-\theta\right)=\cot \theta$$
$$(iv) \,\cot \left(90^{\circ}-\theta\right)=\tan \theta$$
$$(v) \,\sec \left(90^{\circ}-\theta\right)={cosec} \theta$$
$$(vi)\,{cosec}\left(90^{\circ}-\theta\right)=\sec \theta$$

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