Question

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

Answer 1

For the maximum value of $\frac{1}{\sin ^{2} \theta+3 \sin \theta \cos \theta+5 \cos ^{2} \theta}$
$\sin ^{2} \theta+5 \cos ^{2} \theta+3 \sin \theta \cos \theta$ should be minimum.
$\sin ^{2} \theta+5 \cos ^{2} \theta+3 \sin \theta \cos \theta=\sin ^{2} \theta+\cos ^{2} \theta+4 \cos ^{2} \theta+\frac{3}{2} \cdot 2 \sin \theta \cos \theta$
$=1+4 \cos ^{2} \theta+\frac{3}{2} \sin 2 \theta \left(\because \sin ^{2} A+\cos ^{2} A=1 \& 2 \sin A \cos A=\sin \right. $
$=1+4\left(\frac{1+\cos 2 \theta}{2}\right)+\frac{3}{2} \sin 2 \theta \left(\because \cos ^{2} A=\frac{1+\cos 2 A}{2}\right) $
$=\frac{3}{2} \sin 2 \theta+2(1+\cos 2 \theta)+1 $
$=3+\frac{3}{2} \sin 2 \theta+\cos 2 \theta$
We know that,
$-\sqrt{a^{2}+b^{2}} \leq a \cos \theta+b \sin \theta \leq \sqrt{a^{2}+b^{2}}$
Therefore, minimum value of $\frac{3}{2} \sin 2 \theta+2 \cos 2 \theta+3$
$=-\sqrt{\left(\frac{3}{2}\right)^{2}+2^{2}}+3 $
$=-\frac{5}{2}+3$
$=\frac{1}{2}$
Hence, the maximum value of given function is $2$ .