If $\tan \mathrm{A}\cdot \tan \mathrm{B}=\frac{1}{2}\text{, then }(5-3\cos 2\mathrm{A})(5-3\cos 2\mathrm{B})=\mathrm{\_\_\_\_\_\_\_}$

$$\begin{gathered} \begin{array}{l}\tan \mathrm{A}\cdot \tan \mathrm{B}=\frac{1}{2}\Rightarrow \frac{\tan \mathrm{A}}{1}=\frac{\cot \mathrm{B}}{2}=\mathrm{\lambda }\text{ (say) }\\ \Rightarrow \tan \mathrm{A}=\mathrm{\lambda }\text{ and }\tan \mathrm{B}=\frac{1}{2\mathrm{\lambda }} \\ (5-3\cos 2\mathrm{A})(5-3\cos 2\mathrm{B}) \\ =\left(5-\frac{3(1-{\tan }^{2}A)}{1+{\tan }^{2}A}\right)\left(5-\frac{3(1-{\tan }^{2}B)}{1+{\tan }^{2}B}\right)\\ =\left(\frac{2+8{\tan }^{2}A}{1+{\tan }^{2}A}\right)\left(\frac{2+8{\tan }^{2}B}{1+{\tan }^{2}B}\right) \\ =\left(\frac{2+8{\lambda }^2}{1+{\lambda }^2}\right)\left(\frac{2+{\displaystyle \frac{8}{4{\lambda }^2}}}{1+{\displaystyle \frac{1}{4{\lambda }^2}}}\right)=\frac{2(1+4{\lambda }^2)}{1+{\lambda }^2}\frac{2(1+{\lambda }^2)}{{\displaystyle \frac{1+4{\lambda }^2}{4}}}=16\end{array} \end{gathered}$$