If $\frac{\sin \mathrm{A}}{\sin \mathrm{B}}=\frac{\sqrt{3}}{2}$ and $\frac{\cos \mathrm{A}}{\cos \mathrm{B}}=\frac{\sqrt{5}}{2},0<\mathrm{A},\mathrm{B}<\frac{\mathrm{\pi }}{2}$, then

Given that,

$\begin{array}{l}2\sin \mathrm{A}=\sqrt{3}\sin \mathrm{B}\\ \Rightarrow 2\sqrt{5}\sin \mathrm{A}=\sqrt{15}\sin \mathrm{B}----\left(\mathrm{i}\right)\\ \mathrm{and} 2\cos \mathrm{A}=\sqrt{5}\cos \mathrm{B}\\ \Rightarrow 2\sqrt{3}\cos \mathrm{A}=\sqrt{15}\cos \mathrm{B}----\left(\mathrm{ii}\right)\end{array}$

On squaring and adding Eqs. (i) and (ii), we get

$\begin{array}{l}20{\sin }^2\mathrm{A}+12{\cos }^2\mathrm{A}=15\\ \Rightarrow 8{\sin }^2\mathrm{A}=3\Rightarrow {\sin }^2\mathrm{A}=\frac{3}{8}\\ \Rightarrow {\cos }^2\mathrm{A}=\frac{5}{8}\\ \therefore \frac{{\sin }^2\mathrm{A}}{{\cos }^2\mathrm{A}}=\frac{3}{5}\Rightarrow \tan \mathrm{A}=\sqrt{\frac{3}{5}}\end{array}$