Fred1230
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Returning if I have to show the effort, I came to this:
\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}\cdot\frac{\cos\alpha}{1+\cos\alpha}=\tan\frac{\alpha}{2}.
=
\frac{\sin4\alpha}{\sin^2\alpha+cos^2\alpha+\cos4\alpha}\cdot\frac{(\sin^2\alpha+cos^2\alpha)-2sin^2\alpha}{\sin^2\alpha+cos^2\alpha+\cos2\alpha}\cdot\frac{\cos\alpha}{\sin^2\alpha+cos^2\alpha+\cos\alpha}=\frac{\sin\alpha^2}{\cos2\alpha}.
I don't know how to use substitutions
\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}\cdot\frac{\cos\alpha}{1+\cos\alpha}=\tan\frac{\alpha}{2}.
=
\frac{\sin4\alpha}{\sin^2\alpha+cos^2\alpha+\cos4\alpha}\cdot\frac{(\sin^2\alpha+cos^2\alpha)-2sin^2\alpha}{\sin^2\alpha+cos^2\alpha+\cos2\alpha}\cdot\frac{\cos\alpha}{\sin^2\alpha+cos^2\alpha+\cos\alpha}=\frac{\sin\alpha^2}{\cos2\alpha}.
I don't know how to use substitutions