Question

If in a $∆ABC,\tan A+\tan B+\tan C=0$, then $\cot A\cot B\cot C=$
(a) 6
(b) 1
(c) $\frac{1}{6}$
(d) none of these

Answer 1

(d) none of these

ABC is a triangle.

$$\begin{gathered} \therefore A+B+C=\mathrm{\pi } \\ \Rightarrow A+B=\mathrm{\pi }-C \\ \Rightarrow \tan \left(\mathrm{A}+\mathrm{B}\right)=\tan \left(\mathrm{\pi }-\mathrm{C}\right) \\ \Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B}=-\tan C \\ \Rightarrow \tan A\mathit{+}\tan B\mathit{=}\mathit{-}\tan C+\tan A\mathit{}\tan B\mathit{}\tan C \\ \Rightarrow \tan A\mathit{+}\tan B\mathit{+}\tan C\mathit{}\mathit{=}\tan A\mathit{}\tan B\mathit{}\tan C \\ \Rightarrow 0=\tan A\mathit{}\tan B\mathit{}\tan C[\mathrm{Given}:\tan A\mathit{}\tan B\mathit{}\tan C=0] \\ \Rightarrow \tan A\tan B\tan C=0 \\ \Rightarrow \frac{1}{\tan A\mathit{}\tan B\mathit{}\tan C}=\frac{1}{0} \\ \Rightarrow \cot A\cot B\cot C\to \infty \end{gathered}$$