Answer
\[\tan \theta = \frac{{\sin \,\theta }}{{\cos \theta }}\]
Work Step by Step
\[\begin{gathered}
Definition\,:\,trigonometric\,functions \hfill \\
\hfill \\
Let\,P\,\left( {x,y} \right)\,be\,a\,point\,on\,a\,circle\,\,\,with\,radius\,r\,\,associated \hfill \\
with\,the\,angle\,\,\theta .\, \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\sin \,\theta = \frac{y}{r}\,,\,\cos \theta = \frac{x}{r}\,\,,\,\tan \theta = \frac{y}{x} \hfill \\
\hfill \\
\cot \theta = \frac{x}{y}\,\,,\,\sec \theta = \frac{r}{x}\,\,,\,\csc \,\theta = \frac{r}{y} \hfill \\
\hfill \\
Using\,\,definition \hfill \\
\hfill \\
\tan \theta = \frac{y}{x} = \frac{y}{x} \cdot \frac{{\frac{{\frac{1}{r}}}{1}}}{r} = \frac{{\frac{{\frac{y}{r}}}{x}}}{r} \hfill \\
\hfill \\
or \hfill \\
\hfill \\
\tan \theta = \frac{{\sin \,\theta }}{{\cos \theta }} \hfill \\
\end{gathered} \]
