Se va folosi formula fundamentala a trigonometriei pentru a afla sinus de x.
[tex]sin^2x+cos^2x=1 \\\\ sin^2x+(\frac{5}{13})^2=1 \\\\ sin^2x=1^{(169}-\frac{25}{169} \\\\ sin^2x= \frac{169-25}{169} \\\\ sin^2x=\frac{144}{169} \\\\ sin \ x= \sqrt{ \frac{144}{169}} \\\\ sin \ x= \pm \frac{12}{13}[/tex]
Dar x se afla in primul cadran,unde valorile pentru sinus si cosinus sunt pozitive deci:
[tex]sin \ x \to \frac{12}{13} \\\\\\ tg x =\frac{sin x}{cos x} = \frac{\frac{12}{13}}{\frac{5}{13}}= \frac{12}{\not 13}*\frac{\not13}{5} \longrightarrow \frac{12}{5} \ \ \ c.c.t.d[/tex]
