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

$$\eqalign{ & 2\frac{{{{\cos }^2}x - {{\sec }^2}x}}{{{{\tan }^2}x}} = a + b\cos 2x \cr & \Rightarrow 2\frac{{{{\cos }^2}x - \frac{1}{{{{\cos }^2}x}}}}{{\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}} = a + b\cos 2x \cr & \Rightarrow 2\frac{{{{\cos }^4}x - 1}}{{{{\cos }^2}x}} \times \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}} = a + b\cos 2x \cr & \Rightarrow 2\frac{{\left( {{{\cos }^2}x - 1} \right)\left( {{{\cos }^2}x + 1} \right)}}{{\left( {1 - {{\cos }^2}x} \right)}} = a + b\cos 2x \cr & \Rightarrow \frac{{ - 2\left( {1 - {{\cos }^2}x} \right)\left( {{{\cos }^2}x + 1} \right)}}{{\left( {1 - {{\cos }^2}x} \right)}} = a + b\cos 2x \cr & \Rightarrow - 2{\cos ^2}x - 2 = a + b\cos 2x \cr & \Rightarrow - 2 + 1 - 1 - 2{\cos ^2}x = a + b\cos 2x \cr & \Rightarrow - 3 - \left( {2{{\cos }^2}x - 1} \right) = a + b\cos 2x \cr & \Rightarrow - 3 - \cos 2x = a + b{\cos ^2}x \cr & a = - 3,\,\,b = - 1 \cr} $$