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[tex]\frac{\sin ^2 2 \alpha -4\cos ^2\alpha}{\sin ^2\alpha+4\cos ^2\alpha-4}=[/tex]
[tex]\frac{\sin ^2 2 \alpha -4\cos ^2\alpha}{\sin ^2\alpha+4\cos ^2\alpha-4}=[/tex]
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Verified answer
=(4*sin²α*cos²α-4*cos²α)/(sin²α-4*(1-cos²α))=(-4*cos²α*(1-sin²α))/(sin²α-4*sin²α)=-4*cos⁴α/(-3*sin²α)=4/3*cos⁴α/sin²α.