$\begin{vmatrix}
1& 1 & 0 \\[0.3em]
-\cos\,C & \cot\, A & -\cot\, A \\[0.3em]
-\cot\,B & \cot\,C & \cot\, B \end{vmatrix}$
= $\begin{vmatrix}
1& 1 & 0 \\[0.3em]
-\cos\,C & \cot\, A +\cot \, C & -\cot\, A \\[0.3em]
-\cot\,B & \cot \, B + \cot\,C & \cot\, B \end{vmatrix}$
= $ \cot A \cot B + \cot B \cot C + \cot A \cot B + \cot A \cot C$
= $ \cot A \cot B + (\cot A \cot B + \cot B \cot C + \cot C \cot A)$
Since $A + B + C = 108^\circ$
$\therefore\: \tan (A + B) = \tan (180^\circ - C) = - \tan C $
$\Rightarrow$ $\frac{\tan \,A \,\tan \,B }{1 - \tan \, A \, \tan \, B} = - \, \tan \, C $
$\Rightarrow \: \tan A + \tan B + \tan C = \tan A \tan B \tan C $
$\Rightarrow\:\: \frac{\tan \, A}{\tan\, A\, \tan\, B\, \tan \,C } + \frac{\tan \, B}{\tan\, A\, \tan\, B\, \tan \,C } +\frac {\tan \,C}{\tan\, A\, \tan\, B\, \tan \,C } = 1 $
$\Rightarrow\: \cot B \cot C + \cot C \cot A + \cot A \cot B = 1$
$\therefore$ given value = $1 + \cot A \cot B$