If $\cos \mathrm{\alpha }+\cos \mathrm{\beta }=\mathrm{a},\sin \mathrm{\alpha }+\sin \mathrm{\beta }=\mathrm{b}\text{ and }\mathrm{\alpha }-\mathrm{\beta }=2\mathrm{\theta }\text{, }$then $\frac{\cos 3\mathrm{\theta }}{\cos \mathrm{\theta }}=$

$\begin{array}{l}{\mathrm{a}}^2+{\mathrm{b}}^2=4{\cos }^2\mathrm{\theta }\\ \Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2={\cos }^2\mathrm{\alpha }+{\cos }^2\mathrm{\beta }+2\cos \mathrm{\alpha cos}\mathrm{\beta }\\ +{\sin }^2\mathrm{\alpha }+{\sin }^2\mathrm{\beta }+2\sin \mathrm{\alpha sin}\mathrm{\beta }\\ =2+2\cos (\mathrm{\alpha }-\mathrm{\beta })=2+2\cos 2\mathrm{\theta }(\because \mathrm{\alpha }-\mathrm{\beta }=2\mathrm{\theta })\\ =2(1+\cos 2\mathrm{\theta })=4{\cos }^2\mathrm{\theta }\\ \therefore \frac{\cos 3\mathrm{\theta }}{\cos \mathrm{\theta }}=\frac{4{\cos }^3\mathrm{\theta }-3\cos \mathrm{\theta }}{\cos \mathrm{\theta }}\\ =4{\cos }^2\theta -3={\mathrm{a}}^2+{\mathrm{b}}^2-3\end{array}$