# The correct way of simplifying $\frac{4\cos^22\alpha-4\cos^2\alpha+3\sin^2\alpha}{4\cos^2\bigl(\frac{5\pi}{2}-\alpha\bigr)-\sin^22(\alpha-\pi)}$

Using half angle formula, I wrote it as follows $$\frac{4\cos^22\alpha-4\cos^2\alpha+3\sin^2\alpha}{4\cos^2\bigl(\frac{5\pi}{2}-\alpha\bigr)-\sin^22(\alpha-\pi)}=\frac{4\cos4\alpha-7\cos2\alpha+3}{2-2\cos2\alpha-\frac{1-\cos4\alpha}{2}}.$$ But I can't figure out what to do next. In fact, I tried other ways of expressing it, but again had numbers that couldn't be simplified. According to my book

the answer must be $$\frac{8\cos2\alpha+1}{2\cos2\alpha-2}$$

Any comments related to this problem are appreciated!

It's $$\frac{4(2\cos^2\alpha-1)^2-4\cos^2\alpha+3-3\cos^2\alpha}{4(1-\cos^2\alpha)-4(1-\cos^2\alpha)\cos^2\alpha}$$ or
$$\frac{(1-\cos^2\alpha)(7-16\cos^2\alpha)}{4(1-\cos^2\alpha)-4(1-\cos^2\alpha)\cos^2\alpha}$$ or $$\frac{7-16\cos^2\alpha}{4\sin^2\alpha}$$ or $$\frac{7-8(1+\cos2\alpha}{2(1-\cos2\alpha)}$$ or $$\frac{1+8\cos2\alpha}{2\cos2\alpha-2}$$