In the triangle ABC if $\tan A>0,\tan B>0$ and $\tan A\tan B<1$ then the triangle is

Given $\tan A>0,\tan B>0$ and $\tan A\tan B<1$

But $\tan C=\tan \left(180-\overline{A+B}\right)$

$=-\tan \left(A+B\right)$

$=-\frac{\left(\tan A+\tan B\right)}{1-\tan A\tan B}<0$

$\Rightarrow \tan C<0$

$\therefore \overline{)C}$ is obtuse angle.  Hence triangle ABC is obtuse angled triangle.