Question

If angle C of a triangle ABC be obtuse, then prove that $\tan A\tan B<1$.

Answer 1

Since $A+B=\pi -C$, we have
$\tan (A+B)=\tan (\pi -C)$
or $\frac{\tan A+\tan B}{1-\tan A\tan B}=-\tan C>0$, .$\left(1\right)$
$[\because$ C is obtuse angle implies $\tan C<0]$
But since A and B are each less than $\pi /2$, it follows that $\tan A+\tan B>0$
Hence $\left(1\right)$ will hold if $1-\tan A\tan B>0$
or $\tan A\tan B<1$ as required.