Question

The minimum value of ${\cos }^2\theta -6\sin \theta \cos \theta +{\sin }^2\theta +2$ is $a-\sqrt{10}$ Find a

Answer 1

${\cos }^2\theta -6\sin \theta \cos \theta +{\sin }^2\theta +2$
$\frac{1+\cos 2\theta }{2}-3\sin 2\theta +3\frac{(1-\cos 2\theta )}{2}+2$
$=4-\cos 2\theta -3\sin 2\theta$
Now, $-\cos 2\theta -3\sin 2\theta ϵ[-\sqrt{10},\sqrt{10}]$
$\Rightarrow$ $4-\cos 2\theta -3\sin 2\theta ϵ[4-\sqrt{10},4+\sqrt{10}]$
Ans: 4