Quotient identity of Trigonometric functions sine and cosine

Formula

$\large \dfrac{\sin \theta}{\cos \theta} \,=\, \tan \theta$

Proof

right angled triangle

$\Delta CAB$ is a right angled triangle and angle of this triangle is theta $(\theta)$. Express sine and cosine functions in mathematical form.

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

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

In $\Delta CAB$, the length of opposite side, adjacent side and hypotenuse are represented by $BC$, $AB$ and $AC$ respectively. Now, write the sine and cosine functions in terms of lengths of the sides of the triangle.

$\sin \theta$ $\,=\,$ $\dfrac{BC}{AC}$

$\cos \theta$ $\,=\,$ $\dfrac{AB}{AC}$

Divide the sine function by cosine function to understand the quotient law of sine and cosine functions.

$\dfrac{\sin \theta}{\cos \theta}$ $\,=\,$ $\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}$

$\implies \dfrac{\sin \theta}{\cos \theta}$ $\,=\,$ $\dfrac{BC}{AC} \times \dfrac{AC}{AB}$

$\implies \dfrac{\sin \theta}{\cos \theta}$ $\,=\,$ $\dfrac{BC\times AC}{AC \times AB}$

$\implies \require{cancel} \dfrac{\sin \theta}{\cos \theta}$ $\,=\,$ $\dfrac{BC\times \cancel{AC}}{\cancel{AC} \times AB}$

$\implies \dfrac{\sin \theta}{\cos \theta}$ $\,=\,$ $\dfrac{BC}{AB}$

In this right angled triangle, $BC$ and $AB$ represent the length of opposite side and adjacent side.

$\implies \dfrac{\sin \theta}{\cos \theta}$ $\,=\,$ $\dfrac{Length \, of \, Opposite \, side}{Length \, of \, Adjacent \, side}$

The ratio of length of opposite side to adjacent side is tangent of the associated angle and it is written as $\tan \theta$.

$\therefore \,\,\,\,\,\, \dfrac{\sin \theta}{\cos \theta}$ $\,=\,$ $\tan \theta$

The quotient of sine of angle by cosine of angle is equal to tangent of angle. It is called the quotient rule of sine and cosine. It is used to express the ratio of sine to cosine as tangent and also to expand tangent in terms ratio of sine to cosine.

Verification

Assume, the angle of right angled triangle is $60^\circ$. The values of sine, cosine and tangent functions for the angle $60^\circ$ are $\sin 60^\circ \,=\, \dfrac{\sqrt{3}}{2}$, $\cos 60^\circ \,=\, \dfrac{1}{2}$ and $\tan 60^\circ \,=\, \sqrt{3}$

Calculate the ratio of sine of angle $60^\circ$ to cosine of angle $60^\circ$.

$\dfrac{\sin 60^\circ}{\cos 60^\circ} \,=\, \dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}$

$\implies \dfrac{\sin 60^\circ}{\cos 60^\circ} \,=\, \dfrac{\sqrt{3}}{2} \times \dfrac{2}{1}$

$\implies \dfrac{\sin 60^\circ}{\cos 60^\circ} \,=\, \dfrac{\sqrt{3} \times 2}{2 \times 1}$

$\implies \require{cancel} \dfrac{\sin 60^\circ}{\cos 60^\circ} \,=\, \dfrac{\sqrt{3} \times \cancel{2}}{\cancel{2}}$

$\implies \dfrac{\sin 60^\circ}{\cos 60^\circ} \,=\, \sqrt{3}$

The value of $\tan 60^\circ$ is equal to $\sqrt{3}$.

$\therefore \,\,\,\,\,\, \dfrac{\sin 60^\circ}{\cos 60^\circ} \,=\, \tan 60^\circ$

It represents the quotient relation of sine and cosine functions with tangent function.

Save (or) Share
Follow us
Email subscription
Copyright © 2012 - 2017 Math Doubts, All Rights Reserved