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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