Wiemy, że:

$\mathrm{tg}\alpha =\frac{2\sqrt{5}}{5}$

$\alpha$ - kąt ostry

Przekształcając tangensa otrzymamy:

$\frac{\sin \alpha }{\cos \alpha }=\frac{2\sqrt{5}}{5}$

$5\sin \alpha =2\sqrt{5}\cos \alpha$

$\sin \alpha =\frac{2\sqrt{5}}{5}\cos \alpha$

Z jedynki trygonometrycznej mamy:

${\sin }^2\alpha +{\cos }^2\alpha =1$

${\left(\frac{2\sqrt{5}}{5}\cos \alpha \right)}^2+{\cos }^2\alpha =1$

$\frac{4\cdot 5}{25}{\cos }^2\alpha +\frac{25}{25}{\cos }^2\alpha =1$

$\frac{45}{25}{\cos }^2\alpha =1$

${\cos }^2\alpha =\frac{25}{45}$

${\cos }^{}\alpha =\frac{5}{3\sqrt{5}}\cdot \frac{\sqrt{5}}{\sqrt{5}}=\frac{5\sqrt{5}}{3\cdot 5}=\frac{\sqrt{5}}{3}$

$\sin \alpha =\frac{2\sqrt{5}}{5}\cdot \frac{\sqrt{5}}{3}=\frac{2\cdot 5}{5\cdot 3}=\frac{2}{3}$

Dodatkowo zauważmy, że:

$\frac{\cos \alpha }{\sin \alpha }=\frac{1}{\mathrm{tg}\alpha }$

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a)

Obliczamy:

$\frac{6\sin \alpha -8{\cos }^2\alpha }{{\sin }^2\alpha }=\frac{6\sin \alpha }{{\sin }^2\alpha }-\frac{8{\cos }^2\alpha }{{\sin }^2\alpha }=$

$=\frac{6}{\sin \alpha }-8\cdot {\left(\frac{1}{\mathrm{tg}\alpha }\right)}^2=\frac{6}{\frac{2}{3}}-8\cdot {\left(\frac{5}{2\sqrt{5}}\right)}^2=$

$=6\cdot \frac{3}{2}-8\cdot \frac{25}{4\cdot 5}=9-8\cdot \frac{5}{4}=9-10=-1$

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b)

Obliczamy:

$\frac{{\sin }^4\alpha -2{\sin }^2\alpha +1}{1-{\sin }^2\alpha }=\frac{{\left({\sin }^2\alpha -1\right)}^2}{{\cos }^2\alpha }=$

$=\frac{{\left({\cos }^2\alpha \right)}^2}{{\cos }^2\alpha }=\frac{{\cos }^4\alpha }{{\cos }^2\alpha }={\cos }^2\alpha ={\left(\frac{\sqrt{5}}{3}\right)}^2=\frac{5}{9}$