$\begin{array}{l}2\cos \mathrm{\theta }+2\sqrt{2}=3\sec \mathrm{\theta }\\ \therefore 2{\cos }^2\mathrm{\theta }+2\sqrt{2}\cos \mathrm{\theta }-3=0\\ \cos \mathrm{\theta }=\frac{-2\sqrt{2}\pm \sqrt{32}}{4}=\frac{-2\sqrt{2}\pm 4\sqrt{2}}{4}\\ \therefore \cos \mathrm{\theta }=\frac{1}{\sqrt{2}}\text{ or }\cos \mathrm{\theta }=-\frac{3}{\sqrt{2}}\text{ (rejected) }\\ \therefore \mathrm{\theta }=\frac{\mathrm{\pi }}{4}\text{ or }\frac{7\mathrm{\pi }}{2}\Rightarrow \sin \mathrm{\theta }=-\frac{1}{\sqrt{2}};\\ \cot \mathrm{\theta }=-1;\tan \mathrm{\theta }=1\end{array}$