trigonometry - Trouble proving the trigonometric identity $frac{1-2sin(x)}{sec(x)}=frac{cos(3x)}{1+2sin(x)}$

I have become stuck while solving a trig identity. It is:

$$\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}$$

I have simplified the left side as far as I can:

$$\begin{align} \frac{1-2\sin(x)}{\sec(x)} &=\frac{1-2\sin(x)}{1/\cos(x)}=(1-2\sin(x))\cos(x)\\ &=\cos(x)-2\sin(x)\cos(x)=\cos(x)-\sin(2x) \end{align}$$

However, I'm not sure what to do on the right side. I know I can use a compound angle formula to break $\cos(3x)$ into $\cos(2x)\cos(x)-\sin(2x)\sin(x)$; however, I do not know where to go after that. My main problem is with the denominator of the right side, I can't figure out how to get rid of it, either by multiplying, or by using a trig identity. Any help in solving this identity would be greatly appreciated!

Answer

We have that for $\cos x\neq 0$ and $\sin x \neq -\frac12$

$$\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}\iff(1-2\sin(x))(1+2\sin(x))=\frac{\cos (3x)}{\cos x}$$

then recall that $\cos (3x)=4\cos^3x-3\cos x$

$$\iff1-4\sin^2(x)=\frac{4\cos^3x-3\cos x}{\cos x}\iff1-4\sin^2(x)=4\cos^2x-3$$

$$\iff4=4(\cos^2x+\sin^2x)\iff4=4$$