Zadanie 20. (0–1)

$\mathrm{tg}\alpha =\frac{4}{3}$
Korzystamy z tożsamości $\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }$.
$\frac{\sin \alpha }{\cos \alpha }=\frac{4}{3}|\cdot \cos \alpha$
$\sin \alpha =\frac{4}{3}\cos \alpha$

Korzystamy z tożsamości ${\sin }^2\alpha +{\cos }^2\alpha =1$.
${\left(\frac{4}{3}\cos \alpha \right)}^2+{\cos }^2\alpha =1$
$\frac{16}{9}{\cos }^2\alpha +{\cos }^2\alpha =1$
$\frac{25}{9}{\cos }^2\alpha =1|:\frac{25}{9}$
${\cos }^2\alpha =\frac{9}{25}$   i   $\cos \alpha >0$, bo $\alpha$ to kąt ostry
$\cos \alpha =\frac{3}{5}$

$\sin \alpha =\frac{4}{3}\cos \alpha =\frac{4}{3}\cdot \frac{3}{5}=\frac{4}{5}$

$\sin \alpha -\cos \alpha =\frac{4}{5}-\frac{3}{5}=\frac{1}{5}$

Odp.: B