If $\frac{\sin A}{\sin B}=\frac{\sqrt{3}}{2}$ and $\frac{\cos A}{\cos B}=\frac{\sqrt{5}}{2}0<A,B<\pi /2$ then $\tan A+\tan B$ is equal to

From the given relation we have

$\Rightarrow \frac{\tan A}{\sqrt{3}}=\frac{\tan B}{\sqrt{5}}=k\left(\text{ say }\right),(\text{ clearly }k>0)$

Also $2\sin A=\sqrt{3}\sin B$.

$\begin{array}{l}\Rightarrow \frac{2\tan A}{\sqrt{1+{\tan }^{2}A}}=\frac{\sqrt{3}\tan B}{\sqrt{1+{\tan }^{2}B}}\Rightarrow \frac{2\sqrt{3}k}{\sqrt{1+3k^2}}=\frac{\sqrt{3}\times \sqrt{5}k}{\sqrt{1+5k^2}}\\ \Rightarrow 4\left(1+5k^2\right)=5\left(1+3k^2\right)\\ \Rightarrow k^2=1/5\Rightarrow k=1/\sqrt{5}\end{array}$

so that $\tan A=\frac{\sqrt{3}}{\sqrt{5}},\tan B=1\Rightarrow \tan A+\tan B=\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$.