Question:
If in a $\Delta A B C, \tan A+\tan B+\tan C=0$, then $\cot A \cot B \cot C=$
(a) 6
(b) 1
(c) $\frac{1}{6}$
(d) none of these
Solution:
(d) none of these
$A B C$ is a triangle.
$\therefore A+B+C=\pi$
$\Rightarrow A+B=\pi-C$
$\Rightarrow \tan (\mathrm{A}+\mathrm{B})=\tan (\pi-\mathrm{C})$
$\Rightarrow \frac{\tan A+\tan B}{1-\tan A \tan B}=-\tan C$
$\Rightarrow \tan A+\tan B=-\tan C+\tan A \tan B \tan C$
$\Rightarrow \tan A+\tan B+\tan C=\tan A \tan B \tan C$
$\Rightarrow 0=\tan A \tan B \tan C \quad[$ Given $: \tan A \tan B \tan C=0]$
$\Rightarrow \tan A \tan B \tan C=0$
$\Rightarrow \frac{1}{\tan A \tan B \tan C}=\frac{1}{0}$
$\Rightarrow \cot A \cot B \cot C \rightarrow \infty$