Przyjmijmy oznaczenia jak na poniższym rysunku. 

   

$a>0,b>0$  

$\beta <{90}^{\circ }$  

$\cos \beta =\frac{\sqrt{2}}{10}$   

Obliczamy ile wynosi sinus kąta  $\beta$  . 

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

${\sin }^2\beta +{\left(\frac{\sqrt{2}}{10}\right)}^2=1$

${\sin }^2\beta +\frac{2}{100}=1$    

${\sin }^2\beta +\frac{1}{50}=1\mid -\frac{1}{50}$

${\sin }^2\beta =\frac{49}{50}$   

$\sin \beta =\sqrt{\frac{49}{50}}=\frac{7}{\sqrt{25\cdot 2}}=\frac{7}{5\sqrt{2}}=\frac{7}{5\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{7\sqrt{2}}{10}$  

Obliczamy ile wynosi tangens kąta  $\beta$  . 

$\mathrm{tg}\beta =\frac{\sin \beta }{\cos \beta }=\frac{\frac{7\sqrt{2}}{10}}{\frac{\sqrt{2}}{10}}=\frac{7{\sout{\sqrt{2}}}^1}{10}\cdot \frac{10}{{\sout{\sqrt{2}}}^1}=\frac{7}{{\sout{10}}^1}\cdot {\sout{10}}^1=7$  

Zauważmy, że:

$\mathrm{tg}\beta =\frac{b}{a}$  

$\frac{b}{a}=7\mid \cdot a$  

$b=7a$  

Obliczamy ile wynosi wartość podanego wyrażenia. 

$\frac{a}{a+b}-\frac{b^2}{a^2-b^2}=\frac{a}{a+7a}-\frac{{\left(7a\right)}^2}{a^2-{\left(7a\right)}^2}=\frac{{\sout{a}}^1}{8{\sout{a}}^1}-\frac{49a^2}{a^2-49a^2}=$   

$=\frac{1}{8}-\frac{49{\sout{a^2}}^1}{-48{\sout{a^2}}^1}=\frac{1}{8}+\frac{49}{48}=\frac{6}{48}+\frac{49}{48}=\frac{55}{48}$