The maximum value of the expression $\frac{1}{{\sin }^2\mathrm{\theta }+3\sin \mathrm{\theta cos}\mathrm{\theta }+5{\cos }^2\mathrm{\theta }}\text{ is }$

$\begin{array}{l}\text{ Let }\mathrm{f}\left(\mathrm{\theta }\right)=\frac{1}{{\sin }^2\mathrm{\theta }+3\sin \mathrm{\theta cos}\mathrm{\theta }+5{\cos }^2\mathrm{\theta }}\\ \begin{array}{rl}\text{ Again let }\mathrm{g}\left(\mathrm{\theta }\right)& ={\sin }^2\mathrm{\theta }+3\sin \mathrm{\theta cos}\mathrm{\theta }+5{\cos }^2\mathrm{\theta }\\ & =\frac{1-\cos 2\mathrm{\theta }}{2}+5\left(\frac{1+\cos 2\mathrm{\theta }}{2}\right)+\frac{3}{2}\sin 2\mathrm{\theta }\\ & =3+2\cos 2\mathrm{\theta }+\frac{3}{2}\sin 2\mathrm{\theta }\end{array}\\ \therefore \mathrm{g}(\mathrm{\theta }{)}_{min}=-\sqrt{4+\frac{9}{4}}=3-\frac{5}{2}=\frac{1}{2}\\ \therefore \mathrm{f}\left(\mathrm{\theta }\right)=\frac{1}{\mathrm{g}(\mathrm{\theta }{)}_{min}}=2\end{array}$