${\sin }^2\beta ={\left(\sin \left({90}^o-\alpha \right)\right)}^2={\left(\cos \alpha \right)}^2={\cos }^2\alpha$  

 

$\frac{1}{\mathrm{tg}\beta }=\frac{1}{\mathrm{tg}\left({90}^o-\alpha \right)}=\mathrm{tg}\alpha$ $\frac{1}{\mathrm{tg}\beta }=\frac{1}{\mathrm{tg}\left({90}^o-\alpha \right)}=\mathrm{tg}\alpha$  

 

A więc:

$\left({\sin }^2\alpha +{\cos }^2\alpha \right)\left(\mathrm{tg}\alpha \cdot \mathrm{tg}\alpha -\sin \alpha \cos \beta \right)={\mathrm{tg}}^2\alpha -\sin \alpha \cos \beta$  

 

${\left(\mathrm{tg}{60}^o\right)}^2={\left(\sqrt{3}\right)}^2=3$  

$\sin {60}^o=\frac{\sqrt{3}}{2}$  

$\cos {30}^o=\frac{\sqrt{3}}{2}$  

 

Stąd:

$=3-\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}=3-\frac{3}{4}=2\frac{1}{4}$