Zakladam ze jak w poprzednich zadaniach : oblicz tangens
[tex]sin^2\alpha+cos^2\alpha=1\\cos^2\alpha=1-sin^2\alpha\\\\9sin^2\alpha-(1-sin^2\alpha)=1\\9sin^2\alpha-1+sin^2\alpha=1\\10sin^2\alpha-1=1 /+1\\10sin^2\alpha=2 /:10\\sin^2\alpha=\frac2{10}=\frac15\\sin\alpha=\sqrt{\frac15}=\frac1{\sqrt5}=\frac{\sqrt5}5\\\\cos^2\alpha=1-\frac15\\cos^2\alpha=\frac45\\\\cos\alpha=\sqrt{\frac45}=\frac{2}{\sqrt5}=\frac{2\sqrt5}5\\\\tg\alpha=\frac{\sqrt{5}}5:\frac{2\sqrt5}5=\frac{\sqrt5}5*\frac5{2\sqrt5}=\frac{\sqrt5}{2\sqrt5}=\frac12[/tex]
