Convert the given complex number in polar form:

$\sqrt{3}+i$

Let $r \cos \theta=\sqrt{3}$ and $r \sin \theta=1$

On squaring and adding, we obtain

$r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=(\sqrt{3})^{2}+1^{2}$

$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=3+1$

$\Rightarrow r^{2}=4$

$\Rightarrow r=\sqrt{4}=2 \quad[$ Conventionally, $r>0]$

$\therefore 2 \cos \theta=\sqrt{3}$ and $2 \sin \theta=1$

$\Rightarrow \cos \theta=\frac{\sqrt{3}}{2}$ and $\sin \theta=\frac{1}{2}$

$\therefore \theta=\frac{\pi}{6}$ [As $\theta$ lies in the I quadrant]

$\therefore \sqrt{3}+i=r \cos \theta+i r \sin \theta=2 \cos \frac{\pi}{6}+i 2 \sin \frac{\pi}{6}=2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$

This is the required polar form.