Zakładamy, że ${\cos }^{2}2\alpha -{\sin }^{2}2\alpha \ne 0$.

$\displaystyle L=\frac{1-8\sin^2\alpha\cdot\cos^2\alpha}{\cos^22\alpha-\sin^22\alpha}=\frac{1-2\cdot(2\sin\alpha\cdot\cos\alpha)^2}{\cos4\alpha}=$

$\displaystyle=\frac{1-2\sin^22\alpha}{\cos4\alpha}=\frac{\cos4\alpha}{\cos4\alpha}=1=P$

Założenia: ${\displaystyle \frac{1}{\mathrm{tg}2\alpha }-\sin 4\alpha \ne 0}$, $\sin 2\alpha \ne 0$, $\cos 2\alpha \ne 0$.

$\displaystyle L=\frac{\cos^22\alpha-4\sin^2\alpha\cdot\cos^2\alpha}{\frac1{\operatorname{tg}2\alpha}-\sin4\alpha}=\frac{\cos^22\alpha-\sin^22\alpha}{\frac{\cos2\alpha}{\sin2\alpha}-2\sin2\alpha\cdot\cos2\alpha}=$

$\displaystyle=\frac{\cos4\alpha}{\cos2\alpha\cdot\frac{1-2\sin^22\alpha}{\sin2\alpha}}=\frac{\sin2\alpha\cdot\cos4\alpha}{\cos2\alpha\cdot\cos4\alpha}=\operatorname{tg}2\alpha= P$.