Odpowiedź i szczegółowe wyjaśnienie:
[tex](1-cos\alpha)(\dfrac{1}{sin\alpha}+\dfrac{1}{tg\alpha})=sin\alpha[/tex]
Założenia:
sin, tg ≠ 0
[tex]L=(1-cos\alpha)(\dfrac{1}{sin\alpha}+\dfrac{1}{tg\alpha})\\\\P=sin\alpha\\\\\\L=(1-cos\alpha)(\dfrac{1}{sin\alpha}+\dfrac{1}{\frac{sin\alpha}{cos\alpha}})\\\\\\L=(1-cos\alpha)(\dfrac{1}{sin\alpha}+\dfrac{cos\alpha}{sin\alpha})\\\\\\L=(1-cos\alpha)(\dfrac{1+cos\alpha}{sin\alpha})\\\\\\L=\dfrac{(1-cos\alpha)(1+cos\alpha)}{sin\alpha}\\\\\\L=\dfrac{1-cos^2\alpha}{sin\alpha}\\\\\\L=\dfrac{sin^2\alpha}{sin\alpha}\\\\\\L=sin\alpha\\\\L=P[/tex]
