Question

If $A+B+C={90}^{\circ }$,
then the value of
$\frac{\cot A+\cot B+\cot C}{\cot A\cot B\cot C}$
is equal to

Answer 1

Given, $A+B+C={90}^{\circ }$

$\Rightarrow A+B={90}^{\circ }-C$

$\Rightarrow \cot (A+B)=\cot ({90}^{\circ }-C)$

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

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

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

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

Hence the correct answer is Option A.