If 2 ^2x - ^2x ^2x = a + b 2x, then a, b = ?

Answer: Option C

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