In $\mathrm{△}\mathrm{ABC},\tan \mathrm{B}+\tan \mathrm{C}=5\text{ and }\tan \mathrm{Atan}\mathrm{C}=3$then

$\tan \mathrm{A}+\tan \mathrm{B}+\tan \mathrm{C}=\tan \mathrm{Atan}\mathrm{Btan}\mathrm{C}$
$\begin{array}{l}\Rightarrow \tan \mathrm{A}+5=3\tan \mathrm{B}\\ \Rightarrow 5+\tan \mathrm{A}=3(5-\tan \mathrm{C})\\ \Rightarrow 5+\tan \mathrm{A}=15-\frac{9}{\tan \mathrm{A}}\\ \Rightarrow {\tan }^2\mathrm{A}-10\tan \mathrm{A}+9=0\\ \Rightarrow \tan \mathrm{A}=1\text{ or }\mathrm{A}=9\end{array}$
$\Rightarrow \tan B$and tanC are 2, 3 or $\frac{14}{3},\frac{1}{3}$respectively
$\Rightarrow \mathrm{△}ABC$is always on acute angled triangle and sum of all possible values of tanA is 10.