Answer
The polar coordinates of $\left( -\sqrt{3},-1 \right)$ are $\left( 2,\frac{7\pi }{6} \right)$.
Work Step by Step
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 $\left( 2,\frac{7\pi }{6} \right)$.
