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In an acute angled triangle, $\cot B \cot C+\cot A \cot C+\cot A \cot B$ is equal to
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$1$
$\because$ In $\triangle A B C$
$A+B+C=180^{\circ}$
$\begin{array}{ll}\Rightarrow & A+B=180^{\circ}-C \\ \Rightarrow & \cot (A+B)=\cot \left(180^{\circ}-C\right) \\ \Rightarrow & \frac{\cot A \cot B-1}{\cot B+\cot A}=-\cot C \\ \Rightarrow & \cot A \cot B+\cot B \cot C+\cot C \cot A=1\end{array}$
$A+B+C=180^{\circ}$
$\begin{array}{ll}\Rightarrow & A+B=180^{\circ}-C \\ \Rightarrow & \cot (A+B)=\cot \left(180^{\circ}-C\right) \\ \Rightarrow & \frac{\cot A \cot B-1}{\cot B+\cot A}=-\cot C \\ \Rightarrow & \cot A \cot B+\cot B \cot C+\cot C \cot A=1\end{array}$
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