In a $\triangle A B C$, suppose none on the angles are multiples of $\frac{\pi}{2}$, then what is the value…
In a $\triangle A B C$, suppose none on the angles are multiples of $\frac{\pi}{2}$, then what is the value $\cot A \cot B+\cot B \cot C+\cot A \cot C$ ?
$\infty$
1
-1
0
Solution
$\begin{aligned} & \text { Given, } A+B+C=\pi \text { and } A, B, C \neq \frac{n \pi}{2} \\ & \cot (B+C)=\frac{\cot B \cot C-1}{\cot B+\cot C} \\ & \Rightarrow \quad \cot (\pi-A)=\frac{\cot B \cdot \cot C-1}{\cot B+\cot C} \\ & \Rightarrow \quad-\cot A(\cot B+\cot C)=\cot B \cdot \cot C-1 \\ & \Rightarrow \quad \cot A \cot B+\cot B \cot C+\cot C \cot A=1\end{aligned}$