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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!

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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 $$\frac{1+8\cos2\alpha}{2\cos2\alpha-2}$$

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With the help of Michael Rozenberg, I solved it, by using trig functions of half angles: \begin{align} \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+4\cos4\alpha-4-4\cos2\alpha+3-3\cos2\alpha}{2(4\sin^2\alpha-\sin^22\alpha)}&=\\ \frac{4(2\cos^22\alpha-1)-7\cos2\alpha+3}{2(4\sin^2\alpha-\sin^22\alpha)}&=\\ \frac{8\cos^22\alpha-7\cos2\alpha-1}{2(4\sin^2\alpha-\sin^22\alpha)}&=\\ \frac{8\cos^22\alpha-8\cos2\alpha+\cos2\alpha-1}{2(4\sin^2\alpha-\sin^22\alpha)}&=\\ \frac{8\cos2\alpha(\cos2\alpha-1)+(\cos2\alpha-1)}{2(4\sin^2\alpha-\sin^22\alpha)}&=\\ \frac{\cos2\alpha-1(8\cos2\alpha+1)}{2(4\sin^2\alpha-\sin^22\alpha)}&=\\ \frac{\cos2\alpha-1(8\cos2\alpha+1)}{2}:4\sin^2\alpha-\sin^22\alpha&=\\ \frac{\cos2\alpha-1(8\cos2\alpha+1)}{2}:4\cdot\frac{1-\cos2\alpha}{2}-\frac{1-\cos4\alpha}{2}&=\\ \frac{\cos2\alpha-1(8\cos2\alpha+1)}{2}:\frac{1}{\cos2\alpha-1(\cos2\alpha-1)}&=\\ \frac{8\cos2\alpha+1}{2\cos2\alpha-2} \end{align} I know that it is too lengthy.

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