Korzystając z jedynki trygonometrycznej otrzymujemy:

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

${\sin }^2\alpha +{\left(\frac{\sqrt{5}+1}{4}\right)}^2=1$  

${\sin }^2\alpha +\frac{{\left(\sqrt{5}+1\right)}^2}{4^2}=1$       

${\sin }^2\alpha +\frac{5+2\sqrt{5}+1}{16}=1$  

${\sin }^2\alpha +\frac{6+2\sqrt{5}}{16}=1\mid -\frac{6+2\sqrt{5}}{16}$  

${\sin }^2\alpha =\frac{16}{16}-\frac{6+2\sqrt{5}}{16}$  

${\sin }^2\alpha =\frac{10-2\sqrt{5}}{16}$  

Z treści zadania wiemy, że  $\alpha \in \left(\frac{3}{2}\pi ,2\pi \right),\text{zatem}\sin \alpha <0$  

$\sin \alpha =-\sqrt{\frac{10-2\sqrt{5}}{16}}$  

$\sin \alpha =-\frac{\sqrt{10-2\sqrt{5}}}{4}$  

 

Zatem otrzymujemy:

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

$\mathrm{tg}\alpha =\frac{-\frac{\sqrt{10-2\sqrt{5}}}{4}}{\frac{\sqrt{5}+1}{4}}$  

$\mathrm{tg}\alpha =-\frac{\sqrt{10-2\sqrt{5}}}{4}\cdot \frac{4}{\sqrt{5}+1}$  

$\mathrm{tg}\alpha =-\frac{\sqrt{10-2\sqrt{5}}}{\sqrt{5}+1}$  

Usuwając niewymierność z mianownika otrzymujemy:

$\mathrm{tg}\alpha =-\frac{\sqrt{10-2\sqrt{5}}\cdot \left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}$  

$\mathrm{tg}\alpha =-\frac{\sqrt{10-2\sqrt{5}}\cdot \sqrt{{\left(\sqrt{5}-1\right)}^2}}{{\sqrt{5}}^2-1^2}$  

$\mathrm{tg}\alpha =-\frac{\sqrt{10-2\sqrt{5}}\cdot \sqrt{5-2\sqrt{5}+1}}{5-1}$  

$\mathrm{tg}\alpha =-\frac{\sqrt{2}\cdot \sqrt{5-\sqrt{5}}\cdot \sqrt{6-2\sqrt{5}}}{4}$  

$\mathrm{tg}\alpha =-\frac{\sqrt{2}\cdot \sqrt{5-\sqrt{5}}\cdot \sqrt{2}\cdot \sqrt{3-\sqrt{5}}}{4}$  

$\mathrm{tg}\alpha =-\frac{\sqrt{4}\cdot \sqrt{\left(5-\sqrt{5}\right)\left(3-\sqrt{5}\right)}}{4}$  

$\mathrm{tg}\alpha =-\frac{2\cdot \sqrt{15-5\sqrt{5}-3\sqrt{5}+5}}{4}$  

$\mathrm{tg}\alpha =-\frac{\sqrt{20-8\sqrt{5}}}{2}$  

$\mathrm{tg}\alpha =-\frac{\sqrt{4}\cdot \sqrt{5-2\sqrt{5}}}{2}$  

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

 

Odp. A