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
Solution:
(c) $\frac{1}{6}$
In triangle ABC,
$A+B+C=\pi$
We know that $\tan (A+B+C)=\frac{\tan \mathrm{A}+\tan B+\tan C-\tan A \tan B \tan C}{1-\tan A \tan B-\tan B \tan C-\tan C \tan A}$
and $\tan \pi=0$
$\therefore \tan A+\tan B+\tan C-\tan A \tan B \tan C=0$
$\tan A+\tan B+\tan C=\tan A \tan B \tan C$
If tan A+tan B+tan C =6,
tan A tan B tan C =6
$\Rightarrow \frac{1}{\tan A \tan B \tan C}=\frac{1}{6}$
$\Rightarrow \cot A \cot B \cot C=\frac{1}{6}$
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