If ${\cot }^2{\mathrm{Acot}}^2\mathrm{B}=3$ ,then the value of  $(2-\cos 2\mathrm{A})(2-\cos 2\mathrm{B})$ is

$\begin{array}{l}(2-\cos 2\mathrm{A})(2-\cos 2\mathrm{B})\\ =\left(1+2{\sin }^2\mathrm{A}\right)\left(1+2{\sin }^2\mathrm{B}\right)\\ =1+2\left({\sin }^2\mathrm{A}+{\sin }^2\mathrm{B}\right)+4{\sin }^2{\mathrm{Asin}}^2\mathrm{B}----\left(\mathrm{i}\right)\end{array}$

$\begin{array}{l}\mathrm{now}, {\cot }^2{\mathrm{Acot}}^2\mathrm{B}=3\\ \Rightarrow {\cos }^2{\mathrm{Acos}}^2\mathrm{B}=3{\sin }^2{\mathrm{Asin}}^2\mathrm{B}\\ \Rightarrow \left(1-{\sin }^2\mathrm{A}\right)\left(1-{\sin }^2\mathrm{B}\right)=3{\sin }^2{\mathrm{Asin}}^2\mathrm{B}\\ \Rightarrow {\sin }^2\mathrm{A}+{\sin }^2\mathrm{B}+2{\sin }^2{\mathrm{Asin}}^2\mathrm{B}=1\end{array}$

From (1), we get 

$(2-\cos 2\mathrm{A})(2-\cos 2\mathrm{B})=1+2=3$