In an acute angled triangle ABC, $\mathrm{cotA} \mathrm{cotB} + \mathrm{cotB} \mathrm{cotC} + \cot \hspace{0.17em}\mathrm{C} \mathrm{cotA}=\dots \dots$

$$\begin{gathered} \mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{\pi } \\ \cot (\mathrm{A}+\mathrm{B})=\cot (\mathrm{\pi }-\mathrm{c}) \\ \frac{\cot \mathrm{A} \cot \mathrm{B}-1}{\cot \mathrm{A}+\cot \mathrm{B}}=-\cot \mathrm{C} \\ \cot \mathrm{A}+\cot \mathrm{B}-1=\cot \mathrm{C} \cot \mathrm{A}-\cot \mathrm{B} \cot \mathrm{C} \\ \cot \mathrm{A} \cot \mathrm{B}+\cot \mathrm{B} \cot \mathrm{C}+\cot \mathrm{C} \cot \mathrm{A}=1 \end{gathered}$$