Question

If $A+B+C={90}^o$, then $\frac{\cot A+\cot B+\cot C}{\cot A\cot B\cot C}$ is equal to:

• A. $1$
• B. $\cot A\cos B\cot C$
• C. $-1$
• D. $0$

Answer 1

The correct option is A $1$

$A+B={90}^o-C$

Take $\cot$ on both sides, we get

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

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

Cross-Multiply both sides, we get

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

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

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