Avem formulele: [tex]sinx= \frac{tg^2x}{ \sqrt{1+tg^2x} }= \frac{4}{ \sqrt{5} },si, cosx= \frac{1}{ \sqrt{1+tg^2x} }= \frac{1}{ \sqrt{5} } [/tex], daca tgx=2. Deci avem: [tex] \frac{ \frac{4}{5} }{ 4\frac{2}{5}+3 }= \frac{ \frac{4}{5} }{ \frac{8}{5}+ \frac{15}{5} }= \frac{4}{23} [/tex]
