Rozwiązanie:
[tex]sin^{2}\alpha =\frac{4}{5} \\\alpha \in (0^{o},90^{o})[/tex]
Stąd:
[tex]sin\alpha =\frac{2}{\sqrt{5} } \\cos^{2}\alpha =1-sin^{2}\alpha =1-\frac{4}{5}=\frac{1}{5}\\cos\alpha =\frac{1}{\sqrt{5} }[/tex]
Dalej mamy:
[tex](tg\alpha +\frac{1}{tg\alpha})^{2}=(\frac{sin\alpha }{cos\alpha } +\frac{cos\alpha }{sin\alpha })^{2} =(\frac{sin^{2}\alpha +cos^{2}\alpha }{sin\alpha cos\alpha } )^{2}=(\frac{1}{sin\alpha cos\alpha } )^{2}=(\frac{1}{\frac{2}{\sqrt{5} } \cdot \frac{1}{\sqrt{5} } } )^{2}=(\frac{5}{2})^{2}=\frac{25}{4} =6\frac{1}{4}=6.25[/tex]
co kończy dowód.
