$\mathrm{\alpha }\text{ and }\mathrm{\beta }$ are the positive acute angles and satisfying equations $5\sin 2\mathrm{\beta }=3\sin 2\mathrm{\alpha }\text{ and }\tan \mathrm{\beta }=3\tan \mathrm{\alpha }$ simultaneously. Then the value of

$\tan \mathrm{\alpha }+\tan \mathrm{\beta }$ is

$\begin{array}{r}5\times \frac{2\tan \mathrm{\beta }}{1+{\tan }^2\mathrm{\beta }}=3\times \frac{2\tan \mathrm{\alpha }}{1+{\tan }^2\mathrm{\alpha }}\\ \text{ or }\frac{5\tan \mathrm{\beta }}{1+{\tan }^2\mathrm{\beta }}=\frac{3\tan \mathrm{\alpha }}{1+{\tan }^2\mathrm{\alpha }}\end{array}$

$\text{ Substituting }\tan \mathrm{\beta }=3\tan \mathrm{\alpha }\text{, we have }$

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{5\times 3\tan \mathrm{\alpha }}{1+9{\tan }^2\mathrm{\alpha }}=\frac{3\tan \mathrm{\alpha }}{1+{\tan }^2\mathrm{\alpha }}\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}5+5{\tan }^2\mathrm{\alpha }=1+9{\tan }^2\mathrm{\alpha }\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4{\tan }^2\mathrm{\alpha }=4\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\tan \mathrm{\alpha }=1,\text{ i.e., }\tan \mathrm{\beta }=3\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\tan \mathrm{\alpha }+\tan \mathrm{\beta }=4\end{array}$