If in a \[∆ ABC, \tan A + \tan B + \tan C = 0\], then
Options
6
1
- \[\frac{1}{6}\]
none of these
Solution
none of these
ABC is a triangle.
\[\therefore A + B + C = \pi\]
\[ \Rightarrow A + B = \pi - C\]
\[ \Rightarrow \tan\left( A + B \right) = \tan\left( \pi - C \right)\]
\[ \Rightarrow \frac{\text{ tan } A + \text{ tan } B}{1 - \text{ tan } A \text{ tan } B} = - \text{ tan } C\]
\[ \Rightarrow \text{ tan } A + \text{ tan } B = - \text{ tan } C + \text{ tan } A \text{ tan } B \text{ tan } C\]
\[ \Rightarrow \text{ tan } A + \text{ tan } B + \text{ tan } C = \text{ tan } A \text{ tan } B \text{ tan } C\]
\[ \Rightarrow 0 = \text{ tan } A \text{ tan } B \text{ tan } C [Given: \text{ tan } A \text{ tan } B \text{ tan } C = 0]\]
\[ \Rightarrow \text{ tan } A \text{ tan } B \text{ tan } C = 0\]
\[ \Rightarrow \frac{1}{\text{ tan } A \text{ tan } B \text{ tan }C} = \frac{1}{0}\]
\[ \Rightarrow \text{ cot } A \text{ cot } B \text{ cot } C \to \infty\]
