Answer
See the explanation below.
Work Step by Step
The distance between two points in rectangular coordinates $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given by
$d=\sqrt{{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}}$
Substitute ${{x}_{1}}={{r}_{1}}\cos {{\theta }_{1}}$, ${{y}_{1}}={{r}_{1}}\sin {{\theta }_{1}}$ and ${{x}_{2}}={{r}_{2}}\cos {{\theta }_{2}}$, ${{y}_{2}}={{r}_{2}}\sin {{\theta }_{2}}$ to get
$\begin{align}
& d=\sqrt{{{\left( {{r}_{2}}\sin {{\theta }_{2}}-{{r}_{1}}\sin {{\theta }_{1}} \right)}^{2}}+{{\left( {{r}_{2}}\cos {{\theta }_{2}}-{{r}_{1}}\sin {{\theta }_{1}} \right)}^{2}}} \\
& =\sqrt{{{r}_{2}}^{2}{{\sin }^{2}}{{\theta }_{2}}+{{r}_{1}}^{2}{{\sin }^{2}}{{\theta }_{1}}-2{{r}_{1}}{{r}_{2}}sin{{\theta }_{1}}\sin {{\theta }_{2}}+{{r}_{2}}^{2}{{\cos }^{2}}{{\theta }_{2}}+{{r}_{1}}^{2}{{\cos }^{2}}{{\theta }_{1}}-2{{r}_{1}}{{r}_{2}}\cos {{\theta }_{1}}\cos {{\theta }_{2}}} \\
& =\sqrt{{{r}_{1}}^{2}\left( {{\sin }^{2}}{{\theta }_{1}}+{{\cos }^{2}}{{\theta }_{1}} \right)+{{r}_{2}}^{2}\left( {{\sin }^{2}}{{\theta }_{2}}+{{\cos }^{2}}{{\theta }_{2}} \right)-2{{r}_{1}}{{r}_{2}}\left( \cos {{\theta }_{1}}\cos {{\theta }_{2}}+\sin {{\theta }_{1}}\sin {{\theta }_{2}} \right)}
\end{align}$
Using ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ and $\operatorname{cosAcosB}+sinAsinB=cos\left( A-B \right)$ we will simplify the above equation as below:
$d=\sqrt{{{r}_{1}}^{2}+{{r}_{2}}^{2}-2{{r}_{1}}{{r}_{2}}\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)}$
Hence, $d=\sqrt{{{r}_{1}}^{2}+{{r}_{2}}^{2}-2{{r}_{1}}{{r}_{2}}\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)}$ has been proved.
