If 2 $\frac{cos^2x-sec^2x}{tan^2x}$ = a + b cos 2x, then a, b= ?

-3, -1

2 \(\frac{cos²x - sec²x}{tan²x}\)

= 2 \(\frac{(cos²x)² - 1}{sin²x}\)

{ we know, sin²x + cos²x = 1 }

= 2 \(\frac{cos4 x - 1}{1 - cos²x}\)

= 2 \(\frac{(cos²x - 1) . (cos²x + 1)}{1 - cos²x}\)

= -  (2cos²x + 2 )

{ using identity, cos²x - 1 = cos 2x }

= - ( 1 + cos 2x  + 2) 

= - cos 2x - 3

And ATQ,

- cos 2x - 3 = a + b cos2x

So, a = -3 and b = -1

Ans :-  -3 , -1