[tex]zad.a\\\\zal.~~1-sin\alpha \neq 0~~\Rightarrow ~~sin\alpha \neq 1~~\Rightarrow ~~\alpha \neq \dfrac{\pi }{2} +2k\pi ~~\land ~~k\in C\\\\\dfrac{cos^{2} \alpha }{1-sin\alpha } =1+sin\alpha \\\\P=1+sin\alpha\\\\L=\dfrac{cos^{2} \alpha }{1-sin\alpha } \cdot \dfrac{1+sin\alpha }{1+sin\alpha } =\dfrac{cos^{2} \alpha \cdot (1+sin\alpha )}{1-sin^{2} \alpha } =\dfrac{cos^{2} \alpha\cdot (1+sin\alpha ) }{cos^{2} \alpha } =1+sin\alpha \\\\L=P~~cbdu.[/tex]
[tex]zad.b\\\\zal.~~\\sin^{2} \alpha \neq 0\\\\\sqrt{sin^{2} \alpha } \neq 0\\\\\mid sin \alpha \mid \neq 0~~~~\Rightarrow ~~\alpha \neq k\pi ~~\land~~k\in C\\\\\dfrac{tg^{2} \alpha }{sin^{2} \alpha } =1+tg^{2} \alpha \\\\L=\dfrac{tg^{2} \alpha }{sin^{2} \alpha } =tg^{2} \alpha \cdot \dfrac{1 }{sin^{2} \alpha } =\dfrac{sin^{2} \alpha }{cos^{2} \alpha } \cdot \dfrac{1 }{sin^{2} \alpha } =\dfrac{1 }{cos^{2} \alpha } \\\\\\[/tex]
[tex]P=1+tg^{2} \alpha=1+\dfrac{sin^{2} \alpha }{cos^{2} \alpha }=\dfrac{cos^{2} \alpha +sin^{2} \alpha }{cos^{2} \alpha } =\dfrac{1}{cos^{2} \alpha } \\\\L=P~~~~cbdu.[/tex]
Korzystałam z : sin²α + cos²α = 1 oraz tgα = sinα/cosα
