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Question
If in ∆ABC, tan A + tan B + tan C = 6, then cot A cot B cot C =
Options
6
1
- \[\frac{1}{6}\]
None of these
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Solution
In triangle ABC,
\[\text{ We know that }\tan\left( A + B + C \right) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan A \tan B - \tan B \tan C - \tan C \tan A}\]
\[\text{ and }\tan \pi = 0 . \]
\[ \therefore \tan A + \tan B + \tan C - \tan A \tan B \tan C = 0\]
\[\tan A + \tan B + \tan C = \tan A \tan B \tan C\]
If tan A+tan B+tan C =6,
tan A tan B tan C =6
\[ \Rightarrow \cot A \cot B \cot C = \frac{1}{6}\]
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