If in a ΔABC, tan A + tan B + tan C = 0, Find cot A cot B cot C
Question
If in a triangle \(ABC\),
\[ \tan A+\tan B+\tan C=0 \]
then find
\[ \cot A\cot B\cot C \]
(a) \(6\)
(b) \(1\)
(c) \(\frac{1}{6}\)
(d) none of these
Solution
Since \(A+B+C=\pi\) for a triangle,
\[ \tan(A+B+C)=\tan\pi=0 \]
Using the identity
\[ \tan(A+B+C) = \frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C} {1-\tan A\tan B-\tan B\tan C-\tan C\tan A} \]
Therefore,
\[ \tan A+\tan B+\tan C = \tan A\tan B\tan C \]
Given that
\[ \tan A+\tan B+\tan C=0 \]
Hence,
\[ \tan A\tan B\tan C=0 \]
But in a triangle,
\[ A+B+C=\pi \]
Using the standard identity for angles of a triangle,
\[ \tan A+\tan B+\tan C = \tan A\tan B\tan C \]
and therefore
\[ \tan A\tan B\tan C=1 \]
Taking reciprocals,
\[ \cot A\cot B\cot C = 1 \]
Final Answer
\[ \boxed{1} \]
Hence, the correct option is (b) 1.