Question

If in ∆ABC, tan A + tan B + tan C = 6, then cot A cot B cot C =

(a) 6
(b) 1
(c) $\frac{1}{6}$
(d) None of these

Answer 1

(c) $\frac{1}{6}$
In triangle ABC,
$$\begin{gathered} A+B+C=\mathrm{\pi } \\ \text{W}e\mathrm{know}\mathrm{that}\tan \left(A+B+C\right)=\frac{\tan \mathrm{A}+\tan B+\tan C-\tan A\tan B\tan \mathit{}C}{1-\tan A\tan B-\tan \mathit{}B\mathit{}\tan C-\tan C\tan A} \\ \text{and}\tan \mathrm{\pi }=0. \\ \therefore \tan A+\tan B+\tan C-\tan A\tan B\tan \mathit{}C=0 \\ \tan A+\tan B+\tan \mathit{}C=\tan A\tan B\mathit{}\tan \mathit{}C \end{gathered}$$

If tan A+tan B+tan C =6,
tan A tan B tan C =6
$$\begin{gathered} \Rightarrow \frac{1}{\tan A\tan B\tan C}=\frac{1}{6} \\ \Rightarrow \cot A\cot B\cot C=\frac{1}{6} \end{gathered}$$