Let

$$f(\theta ) = {1 \over {{{\sin }^2}\theta + 3\sin \theta \cos \theta + 5{{\cos }^2}\theta }}$$

Again let

$$g(\theta ) = {\sin ^2}\theta + 3\sin \theta \cos \theta + 5{\cos ^2}\theta $$

$$ = {{1 - \cos 2\theta } \over 2} + 5\left( {{{1 + \cos 2\theta } \over 2}} \right) + {3 \over 2}\sin 2\theta $$

$$ = 3 + 2\cos 2\theta + {3 \over 2}\sin 2\theta $$

$$\therefore$$ $$g{(\theta )_{\min }} = 3 - \sqrt {4 + {9 \over 4}} = 3 - {5 \over 2} = {1 \over 2}$$

$$\therefore$$ $$f(\theta ) = {1 \over {g{{(\theta )}_{\min }}}} = 2$$