Zauważmy, że:

$\mathrm{tg}\alpha =\frac{3}{4}$  

 

Obliczmy wartość  $\cos \alpha$  : 

$\{\begin{array}{l}{\sin }^2\alpha +{\cos }^2\alpha =1\\ \mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }\end{array}$  

$\{\begin{array}{l}{\sin }^2\alpha +{\cos }^2\alpha =1\\ \frac{3}{4}=\frac{\sin \alpha }{\cos \alpha }\end{array}$  

$\{\begin{array}{l}{\sin }^2\alpha +{\cos }^2\alpha =1\\ \frac{3}{4}\cos \alpha =\sin \alpha \end{array}$  

$\{\begin{array}{l}{\left(\frac{3}{4}\cos \alpha \right)}^2+{\cos }^2\alpha =1\\ \frac{3}{4}\cos \alpha =\sin \alpha \end{array}$  

$\{\begin{array}{l}\frac{9}{16}{\cos }^2\alpha +{\cos }^2\alpha =1\\ \frac{3}{4}\cos \alpha =\sin \alpha \end{array}$  

$\{\begin{array}{l}\frac{25}{16}{\cos }^2\alpha =1\\ \frac{3}{4}\cos \alpha =\sin \alpha \end{array}$  

$\{\begin{array}{l}{\cos }^2\alpha =\frac{16}{25}\\ \frac{3}{4}\cos \alpha =\sin \alpha \end{array}$  

$\{\begin{array}{l}\cos \alpha =\frac{4}{5}\\ \frac{3}{4}\cos \alpha =\sin \alpha \end{array}$  

 

Zatem:

$\frac{2-\cos \alpha }{2+\cos \alpha }=\frac{2-\frac{4}{5}}{2+\frac{4}{5}}=\frac{\frac{6}{5}}{\frac{14}{5}}=\frac{6}{5}\cdot \frac{5}{14}=\frac{6}{14}=\frac{3}{7}$  

 

Odp.: C