Let $\mathrm{f}\left(\mathrm{\theta }\right)=\frac{\cot \mathrm{\theta }}{1+\cot \mathrm{\theta }}\text{ and }\mathrm{\alpha }+\mathrm{\beta }=\frac{5\mathrm{\pi }}{4}$then the value of $\mathrm{f} \left(\mathrm{\alpha }\right) \mathrm{f}\left(\mathrm{\beta }\right)$ is ______

$\begin{array}{l}\cot (\mathrm{\alpha }+\mathrm{\beta })=\cot \frac{5\mathrm{\pi }}{4}=1;\frac{\cot \mathrm{\alpha cot}\mathrm{\beta }-1}{\cot \mathrm{\alpha }+\cot \mathrm{\beta }}=1\\ \Rightarrow \cot \mathrm{\alpha }+\cot \mathrm{\beta }=\cot \mathrm{\alpha cot}\mathrm{\beta }-1\\ \mathrm{f} \left(\mathrm{\alpha }\right) \mathrm{f}\left(\mathrm{\beta }\right)=\frac{\cot \mathrm{\alpha cot}\mathrm{\beta }}{(1+\cot \mathrm{\alpha })(1+\cot \mathrm{\beta })}\\ =\frac{\cot \mathrm{\alpha cot}\mathrm{\beta }}{1+(\cot \mathrm{\alpha }+\cot \mathrm{\beta })+\cot \mathrm{\alpha cot}\mathrm{\beta }}\\ =\frac{\cot \mathrm{\alpha cot}\mathrm{\beta }}{2\cot \mathrm{\alpha cot}\mathrm{\beta }}=\frac{1}{2}\end{array}$