Proving $\frac{\cos 2x}{1+\sin 2 x} = \sec 2 x - \tan 2x$

Question

I want to prove that

$$\frac{\cos 2x}{1+\sin 2 x} = \sec 2 x - \tan 2x$$

I tried $$\frac{1-\sin^2x}{1+2\sin x\cos x}-\frac{1-\cos^2x}{1+2\sin x\cos x}$$ but I couldn't simplify. I start with LHS.

Answer 1

$$\cos y\cdot\cos y=(1-\sin y)(1+\sin y)$$

$$\implies\dfrac{\cos y}{1+\sin y}=\dfrac{1-\sin y}{\cos y}=\dfrac1{\cos y}-\dfrac{\sin y}{\cos y}=?$$

Answer 2

Use that $$1-\sin^2(2x)=\cos^2(2x)$$

Answer 3

$$\left(\frac{1+\sin2x}{\cos2x}\right)^{-1}=(\sec2x+\tan2x)^{-1}=\frac{\sec2x-\tan2x}{\sec^22x-\tan^22x}=\sec2x-\tan2x$$