The quotient relation of sine and cosine functions with tangent function is called the quotient identity of sine and cosine functions with tangent function.

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

Sine function can be divided by the cosine function and the quotient of them represents another trigonometric function tangent. The quotient relation of sine and cosine functions with tangent function is used as basic trigonometric formula in mathematics.

$\Delta BAC$ is a right angled triangle, whose angle is assumed as theta ($\theta$).

Express sine and cosine functions in terms of ratio of the sides of the right angled triangle $BAC$.

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

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

Divide sine function by cosine function and simplify it to obtain quotient of them.

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

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

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

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

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

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

The quotient of sine by cosine function is a ratio of two sides. Actually, the ratio of length of opposite side ($BC$) by length of adjacent side ($AB$) represents tangent function as per the fundamental definition of the tangent function.

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

$\implies \tan{\theta} \,=\, \dfrac{BC}{AB}$

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

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

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