Question

If $A+B+C={180}^o$, 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

Answer: (A)

$1$

Given

$A+B+C={180}^{\circ }$

Applying $\cot$ on both sides

$\cot (A+B+C)=\cot \left(180\right)$

$⟹\frac{\cot A\cot B+\cot B\cot C+\cot B\cot A-1}{\cot A+\cot B+\cot C-\cot A\cot B\cot C}=\frac{1}{0}$

$(\cot (A+B+C)=\frac{\cot A\cot B+\cot B\cot C+\cot B\cot A-1}{\cot A+\cot B+\cot C-\cot A\cot B\cot C})$

$⟹\cot A+\cot B+\cot C-\cot A\cot B\cot C=0$

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

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