Correct Answer: $-3, -1$

Solution : $2 \frac{\cos ^2 x-\sec ^2 x}{\tan ^2 x}$
= $2 \frac{\cos ^2 x-\frac{1}{\cos ^2 x}}{\tan ^2 x}$
= $2 \frac{\cos ^4 x - 1}{\tan ^2 x\cos ^2 x}$
= $2 \frac{(\cos ^2 x - 1)(\cos ^2 x + 1)}{\frac{\sin^2 x}{\cos^2 x}\cos ^2 x}$
We know that $\sin ^2 x + \cos ^2 x=1$
= $2 \frac{(-\sin ^2 x)(\cos ^2 x + 1)}{\sin^2 x}$
So, $2 \frac{\cos ^2 x-\sec ^2 x}{\tan ^2 x} = -2\cos ^2 x-2$
We know that $\cos ^2 x = \frac{(1+\cos 2x)}{2}$
So, $2 \frac{\cos ^2 x-\sec ^2 x}{\tan ^2 x} =-2\frac{(1+\cos 2x)}{2}-2=-3-\cos 2x$
Thus, $a+b\cos2x=-3-\cos 2x$
By comparing we get, $a = -3$ and $b= -1$
Hence, the correct answer is $-3, -1$.