Question : If $2 \frac{\cos ^2 x-\sec ^2 x}{\tan ^2 x}=a+b \cos 2 x$, then $a, b=$?
Option 1: $-\frac{3}{2}, -\frac{1}{2}$
Option 2: $\frac{3}{2}, \frac{1}{2}$
Option 3: $-3, -1$
Option 4: $3,1$
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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 $-3, -1$.
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