If in a Δ ABC,

Question:

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

(a) 6

(b) 1

(c) $\frac{1}{6}$

(d) none of these

Solution:

(d) none of these

$A B C$ is a triangle.

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

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

$\Rightarrow \tan (\mathrm{A}+\mathrm{B})=\tan (\pi-\mathrm{C})$

$\Rightarrow \frac{\tan A+\tan B}{1-\tan A \tan B}=-\tan C$

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

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

$\Rightarrow 0=\tan A \tan B \tan C \quad[$ Given $: \tan A \tan B \tan C=0]$

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

$\Rightarrow \frac{1}{\tan A \tan B \tan C}=\frac{1}{0}$

$\Rightarrow \cot A \cot B \cot C \rightarrow \infty$

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