The relation of ratio of sine to cosine with tangent at an angle is called quotient trigonometric identity of sine and cosine with tangent.

Trigonometric functions sine and cosine mutually have relation in some forms and quotient form is one of them. Sine and cosine forms another trigonometric ratio tangent by their ratio form. Due to formation of this relation by the quotient of ratio of sine to cosine, it is called as a quotient trigonometric identity.

The ratio of sine to cosine actually forms tangent as its quotient and the reason is, the trigonometric ratios sine and cosine both have length of hypotenuse in denominator commonly.

$\Delta CAB$ is a right angled triangle and angle of this triangle is theta $(\theta)$.

$$\sin \theta = \frac{Length \, of \, Opposite \, side}{Length \, of \, Hypotenuse}$$

$$\cos \theta = \frac{Length \, of \, Adjacent \, side}{Length \, of \, Hypotenuse}$$

Divide sine by cosine to get the quotient of them.

$$ \frac{\sin \theta}{\cos \theta} = \frac{\frac{Length \, of \, Opposite \, side}{Length \, of \, Hypotenuse}}{\frac{Length \, of \, Adjacent \, side}{Length \, of \, Hypotenuse}}$$

$$\Rightarrow \frac{\sin \theta}{\cos \theta} = \frac{Length \, of \, Opposite \, side}{Length \, of \, Hypotenuse} \times \frac{Length \, of \, Hypotenuse}{Length \, of \, Adjacent \, side}$$

$$\Rightarrow \frac{\sin \theta}{\cos \theta} = \frac{Length \, of \, Opposite \, side}{Length \, of \, Adjacent \, side} \times \frac{Length \, of \, Hypotenuse}{Length \, of \,Hypotenuse}$$

$$\Rightarrow \frac{\sin \theta}{\cos \theta} = \frac{Length \, of \, Opposite \, side}{Length \, of \, Adjacent \, side}$$

According to $\Delta CAB$

$$\frac{Length \, of \, Opposite \, side}{Length \, of \, Adjacent \, side} = \tan \theta$$

Substitute it in the above expression to get the relation of sine and cosine with tangent.

$$\frac{\sin \theta}{\cos \theta} = \tan \theta$$

It is proved that the value of ratio of sine to cosine is equal to tangent at an angle.

Assume angle of right angled triangle $CAB$ is $60^°$.

Check Left hand side expression by substituting $\theta$ by $60^°$.

$$\frac{\sin \theta}{\cos \theta} = \frac{\sin 60^°}{\cos 60^°}$$

$$\Rightarrow \frac{\sin 60^°}{\cos 60^°} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$$

$$\Rightarrow \frac{\sin 60^°}{\cos 60^°} = \frac{\sqrt{3}}{2}\times\frac{2}{1}$$

$$\Rightarrow \frac{\sin 60^°}{\cos 60^°} = \sqrt{3}$$

Check right hand side expression by doing same.

$\tan \theta = \tan 60^°$

$\Rightarrow \tan 60^° = \sqrt{3}$

The quotient of ratio of sine to cosine at angle $60^°$ is $\sqrt{3}$ and the value of tangent at angle $60^°$ is also equal to $\sqrt{3}$.

Therefore, it is proved that the quotient of ratio of sine to cosine at an angle is equal to tangent at same angle. It is also proved in case of every angle. Hence, it is called as a trigonometric identity.

It is also called as Quotient trigonometric identity due to the formation by the quotient of two trigonometric functions.

Copyright © 2012 - 2017 Math Doubts, All Rights Reserved