## Precalculus (6th Edition) Blitzer

We are given $\theta =\frac{\pi }{3}$. Therefore, \begin{align} & \tan \theta =\tan \frac{\pi }{3} \\ & \tan \theta =\sqrt{3} \end{align} As $x=r\cos \theta \ \text{ and }\ y=r\sin \theta$. \begin{align} & \frac{r\sin \theta }{r\cos \theta \ }=\frac{y}{x} \\ & \tan \theta =\frac{y}{x} \end{align} Also, $\tan \theta =\sqrt{3}$ Therefore, \begin{align} & \frac{y}{x}=\sqrt{3} \\ & y=\sqrt{3}x \end{align} Thus, the provided equation is the equation of line with slope $\sqrt{3}$.