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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.

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  • $\begingroup$ Please use MathJax for formatting mathematics. As it stands, I can't tell whether sin2x means $\sin(2x)$ or $\sin^2x$. $\endgroup$ – Arthur Apr 6 '18 at 9:51
  • $\begingroup$ Since "$2x$" appears everywhere, it's a bit of an unnecessary distraction. Replacing each $2x$ with, say, $y$ gives $$\frac{\cos y}{1+\sin y} = \sec y - \tan y$$ In this form, you see that there's no need to complicate things by invoking the double-angle formulas. $\endgroup$ – Blue Apr 6 '18 at 10:04
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Use that $$1-\sin^2(2x)=\cos^2(2x)$$

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

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

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

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