Maximum Value of Trigonometric Fraction Expression

To find the maximum value of the given expression, let:
$E = \frac{1}{\sin^2 \theta + 3\sin \theta \cos \theta + 5\cos^2 \theta}$

Let the denominator be denoted by $D(\theta)$:
$D(\theta) = \sin^2 \theta + 3\sin \theta \cos \theta + 5\cos^2 \theta$

Using standard trigonometric identities to express $D(\theta)$ in terms of double angles:
$\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$
$\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$
$2\sin \theta \cos \theta = \sin 2\theta$

Substituting these into the expression for $D(\theta)$:
$D(\theta) = \frac{1 - \cos 2\theta}{2} + \frac{3}{2}\sin 2\theta + 5\left(\frac{1 + \cos 2\theta}{2}\right)$
$D(\theta) = \frac{1 + 5}{2} + \frac{3}{2}\sin 2\theta + \left(\frac{5 - 1}{2}\right)\cos 2\theta$
$D(\theta) = 3 + \frac{3}{2}\sin 2\theta + 2\cos 2\theta$

For an expression of the form $A\sin x + B\cos x$, the range of values is given by:
$\left[-\sqrt{A^2 + B^2}, \, \sqrt{A^2 + B^2}\right]$

Here, $A = \frac{3}{2}$ and $B = 2$, so:
$\sqrt{A^2 + B^2} = \sqrt{\left(\frac{3}{2}\right)^2 + 2^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$

Therefore, the minimum value of $D(\theta)$ occurs when $\frac{3}{2}\sin 2\theta + 2\cos 2\theta$ reaches its minimum value of $-\frac{5}{2}$:
$D_{\min} = 3 - \frac{5}{2} = \frac{1}{2}$

Since $D(\theta) > 0$ for all $\theta$, the maximum value of the given expression $E$ corresponds to the reciprocal of the minimum value of the denominator:
$E_{\max} = \frac{1}{D_{\min}} = \frac{1}{\frac{1}{2}} = 2$