Question

If in a $\Delta ABC$, tan A + tan B + tan C = 0, then cot A cot B cot C =

• A. 6
• B. 1
• C. $\frac{1}{6}$
• D. none of these

Answer 1

The correct option is D

none of these

ABC is a triangle.

$\therefore$ A + B + C = $\pi$

$\Rightarrow A+B=\pi -C\Rightarrow tan(A+B)=tan(\pi -C)\Rightarrow \frac{tanA+tanB}{1-tanAtanB}=-tanC$

$\Rightarrow$ tan A + tan B =-tan C + tan A tan B tan C

$\Rightarrow$ tan A + tan B + tan C = tan A tan B tan C

$\Rightarrow$ 0 = tan A tan B tan C = 0

$\Rightarrow$ tan A tan B tan C = 0

$\Rightarrow \frac{1}{tanAtanBtanC}=\frac{1}{0}\Rightarrow cotAcotBcotC\to \infty$