Odpowiedź :
[tex]\frac{sin^2\alpha}{1-cos\alpha}=1+cos\alpha\\\\L=\frac{sin^2\alpha}{1-cos\alpha}=\frac{1-cos^2\alpha}{1-cos\alpha}=\frac{(1-cos\alpha)(1+cos\alpha)}{1-cos\alpha}=1+cos\alpha=P[/tex]
[tex]\frac{sin^2\alpha}{1-cos\alpha}=1+cos\alpha\\\\L=\frac{sin^2\alpha}{1-cos\alpha}=\frac{1-cos^2\alpha}{1-cos\alpha}=\frac{(1-cos\alpha)(1+cos\alpha)}{1-cos\alpha}=1+cos\alpha=P[/tex]