# Proof for Sine by Cosine Quotient identity

According to the quotient of sine by cosine identity, the quotient of sine by cosine is equal to tangent.

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

This quotient trigonometric identity is derived in mathematical form geometrically from a right triangle. Now, it is your turn to learn how to prove the quotient of sine by cosine trigonometric identity from a geometric shape right angled triangle.

$\Delta BAC$ is a right triangle. In this triangle, $\overline{BC}$, $\overline{AB}$ and $\overline{AC}$ are opposite side, adjacent side and hypotenuse respectively. The angle of triangle is taken as theta. ### Define Sine and Cosine functions

According to the definitions of the trigonometric functions, write the sine and cosine functions in ratio form at an angle theta.

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

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

### Divide the Sine by Cosine function

Now, divide the sine function by cosine function to obtain the quotient of them.

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

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

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

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

$\implies$ $\dfrac{\sin{\theta}}{\cos{\theta}}$ $\,=\,$ $\dfrac{BC}{AB} \times 1$

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

### Express the Ratio in Trigonometric function

In the above step, it is derived that the ratio of sine by cosine is equal to the quotient of lengths of opposite side ($BC$) by adjacent side ($AB$).

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

As per trigonometry, the quotient of $BC$ by $AB$ represents the tangent. In this case, the angle of right triangle is theta. Hence, the tan function is written as $\tan{\theta}$.

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

Therefore, it is proved that the quotient of sine function by cosine function is equal to the tangent function mathematically.

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