Question

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

Answer 1

Answer: 2

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.
${\sin }^2\theta +5{\cos }^2\theta +3\sin \theta \cos \theta ={\sin }^2\theta +{\cos }^2\theta +4{\cos }^2\theta +\frac{3}{2}\cdot 2\sin \theta \cos \theta$
$=1+4{\cos }^2\theta +\frac{3}{2}\sin 2\theta (\because {\sin }^{2}A+{\cos }^{2}A=1\&2\sin A\cos A=\sin$
$=1+4\left(\frac{1+\cos 2\theta }{2}\right)+\frac{3}{2}\sin 2\theta \left(\because {\cos }^{2}A=\frac{1+\cos 2A}{2}\right)$
$=\frac{3}{2}\sin 2\theta +2(1+\cos 2\theta )+1$
$=3+\frac{3}{2}\sin 2\theta +\cos 2\theta$
We know that,
$-\sqrt{a^2+b^2}\le a\cos \theta +b\sin \theta \le \sqrt{a^2+b^2}$
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 $2$ .