## Precalculus (6th Edition) Blitzer

The polar coordinates of $\left( -\sqrt{3},-1 \right)$ are $\left( 2,\frac{7\pi }{6} \right)$.
The polar coordinates of the point are $\left( r,\theta \right)$. Now rewrite polar coordinates in terms of rectangular coordinates as below: $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ …… (1) $\tan \theta =\frac{y}{x}$ …… (2) Substituting the values of $x\ \text{ and }\ y$ in (1) and (2), we get \begin{align} & r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\ & =\sqrt{{{\left( -\sqrt{3} \right)}^{2}}+{{\left( -1 \right)}^{2}}} \\ & =\sqrt{3+1}=\sqrt{4} \\ & r=2 \end{align} And, \begin{align} & \tan \theta =\frac{y}{x} \\ & =\frac{-1}{-\sqrt{3}} \\ & \tan \theta =\frac{1}{\sqrt{3}} \end{align} Hence, $\tan \theta =\frac{1}{\sqrt{3}}$ And, $\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$ Also, $\theta$ lies in the third quadrant which means \begin{align} & \theta =\pi +\frac{\pi }{6} \\ & \theta =\frac{7\pi }{6} \end{align} Therefore, the polar coordinates of $\left( -\sqrt{3},-1 \right)$ are $\left( 2,\frac{7\pi }{6} \right)$.