Answer
$d= \sqrt{r_{1}^{2}-2r_{1}r_{2}\cos(\theta_{1}-\theta_{2})+r_{2}^{2}}$.
Work Step by Step
The Cartesian coordinates
for point $(r_{1}, \theta_{1})$ are $(r_{1}\cos\theta_{1}, r_{1}\sin\theta_{1})$
and for $(r_{2}, \theta_{2})$ are $(r_{2}\cos\theta_{2}, r_{2}\sin\theta_{2})$,
We use the distance formula between two points when given in Cartesian coordinates,
$d^{2}=(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$
$= (r_{2}\cos\theta_{2}-r_{1}\cos\theta_{1})^{2}+(r_{2}\sin\theta_{2}-r_{1}\sin\theta_{1})^{2}$
$=r_{2}^{2}\cos^{2}\theta_{2}-2r_{1}r_{2}\cos\theta_{1}\cos\theta_{2}+r_{1}^{2}\cos^{2}\theta_{1})+(r_{2}^{2}\sin^{2}\theta_{2}-2r_{1}r_{2}\sin\theta_{1}\sin\theta_{2}+r_{1}^{2}\sin^{2}\theta_{1})$
$=r_{1}^{2}(\sin^{2}\theta_{1}+\cos^{2}\theta_{1})+r_{2}^{2}(\sin^{2}\theta_{2}+\cos^{2}\theta_{2})-2r_{1}r_{2}(\cos\theta_{1}\cos\theta_{2}+\sin\theta_{1}\sin\theta_{2})$
... we recognize $\sin^{2}A+\cos^{2}A=1$ and $\cos(A-B)=\cos A\cos B+\sin A+\sin B...$
$d^{2}=r_{1}^{2}-2r_{1}r_{2}\cos(\theta_{1}-\theta_{2})+r_{2}^{2}$,
$d= \sqrt{r_{1}^{2}-2r_{1}r_{2}\cos(\theta_{1}-\theta_{2})+r_{2}^{2}}$.
