1-cosα≠0 ∧ 1+cosα≠0 ∧ sin²α≠0
cosα≠1 ∧ cosα≠-1 ∧ sinα≠0
α≠0+2kπ ∧ α≠π+2kπ ∧ α≠0+kπ k∈C
Z.:
α ≠ kπ, k∈C
T.:
[tex]\dfrac1{1-\cos\alpha}+\dfrac1{1+\cos\alpha}=\dfrac{2}{\sin^2\alpha}[/tex]
D.:
[tex]L=\dfrac1{1-\cos\alpha}+\dfrac1{1+\cos\alpha}=\\\\\\=\dfrac{1+\cos\alpha}{(1-\cos\alpha)(1+\cos\alpha)}+\dfrac{1-\cos\alpha}{(1-\cos\alpha)(1+\cos\alpha)}=\\\\\\=\dfrac{1+\cos\alpha+1-\cos\alpha}{1-\cos^2\alpha}=\dfrac{2}{\sin^2\alpha}=P[/tex]
