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