Rozwiązanie:
[tex]1+3ctg^{2}\alpha =\frac{3}{sin^{2}\alpha } -2\\P=\frac{3-2sin^{2}\alpha }{sin^{2}\alpha } =\frac{2-2sin^{2}\alpha +1}{sin^{2}\alpha }=\frac{2(1-sin^{2}\alpha )+1}{sin^{2}\alpha }=\frac{2cos^{2}\alpha +1}{sin^{2}\alpha }= 2ctg^{2}\alpha +\frac{1}{sin^{2}\alpha }\\L=1+3ctg^{2}\alpha =\frac{sin^{2}\alpha +3cos^{2}\alpha }{sin^{2}\alpha }=\frac{1+2cos^{2}\alpha }{sin^{2}\alpha } =\frac{1}{sin^{2}\alpha } +2ctg^{2}\alpha[/tex]
Zatem [tex]L=P[/tex], co kończy dowód.
