If $\tan A+\tan B+\tan C=6$ and $\tan A\tan B=2 $ in $\triangle ABC$, then find the type of triangle. 
In $\triangle ABC$, $\tan A+\tan B+\tan C=6 \\
\tan A\tan B=2
$
Then the triangle is 

$a.)\text{Right-angled isosceles} \\
 b.) \text{Acute-angled isosceles}\\
\color{green}{c.)\text{Obtuse-angled}} \\
 d.)\text{equilateral} $
$\ \ \ $
$\tan A+\tan B+\tan C=6 \\
\tan A\tan B=2 \\
A+B=180-C\\
\dfrac{\tan A+\tan B}{1-\tan A\tan B}=-\tan C\\
\tan A+\tan B=\tan C\\
\tan C=3 \\
\tan A\tan B=2\ \text{and} \ \tan A+\tan B=3  \\
\implies \tan A=2, \tan B=1\ \ \text{or}\ \  \tan A=1, \tan B=2
 $
Now I am stucked, I look for a short and simple way.
I have studied maths upto $12$th grade .
Note:- Calculator is not allowed.

WLOG, $C = \tan^{-1}(3) \approx 72^{\circ}, A = \tan^{-1}(1) = 45^{\circ}, B = \tan^{-1}(2) \approx 63^{\circ}$. This shows that $\triangle ABC$ is scalene !
