(b) $A + B + C = \pi \Rightarrow A + B = \pi – C$

$ \Rightarrow \tan (A + B) = \tan (\pi – C)$

$ \Rightarrow \frac{{\tan A + \tan B}}{{1 – \tan A\tan C}} = \tan (\pi – C)$

$ \Rightarrow \frac{{\tan A + \tan B}}{{1 – \tan A\tan B}} = – \tan C$

Now $C$ is an obtuse angle, hence

$ \Rightarrow \tan C < 0 \Rightarrow – \tan C > 0$

$ \Rightarrow \frac{{\tan A + \tan B}}{{1 – \tan A\tan B}} > 0$

$\Rightarrow 1 – \tan A\tan B > 0$

$(\because A,B$ are acute angles; $\therefore \tan A > 0,\tan B > 0 )$

$ \Rightarrow \tan A\tan B < 1$.