If tan A + tan B + tan C = 6, Find cot A cot B cot C

Solution

In a triangle,

\[ A+B+C=\pi \]

Using the identity:

\[ \tan(A+B+C)=0 \]

Now,

\[ \tan(A+B+C) = \frac{ \tan A+\tan B+\tan C -\tan A\tan B\tan C } { 1-\tan A\tan B-\tan B\tan C-\tan C\tan A } \]

Since

\[ A+B+C=\pi \]

we have

\[ \tan(A+B+C)=\tan\pi=0 \]

Therefore, numerator must be zero:

\[ \tan A+\tan B+\tan C = \tan A\tan B\tan C \]

Given,

\[ \tan A+\tan B+\tan C=6 \]

Hence,

\[ \tan A\tan B\tan C=6 \]

Taking reciprocal,

\[ \cot A\cot B\cot C = \frac{1}{6} \]

Final Answer

\[ \boxed{ \cot A\cot B\cot C=\frac{1}{6} } \]

Correct Option: (c)