Sabendo que \({\mathop{\rm sen}\nolimits} \alpha = \frac{1}{3}\)

Aplicando a Fórmula Fundamental da Trigonometria, vem:

\[\begin{array}{*{20}{l}}{{{\left( {\frac{1}{3}} \right)}^2} + {{\cos }^2}\alpha = 1}& \Leftrightarrow &{{{\cos }^2}\alpha = 1 – \frac{1}{9}}\\{}& \Leftrightarrow &{{{\cos }^2}\alpha = \frac{8}{9}}\end{array}\]

Como \(\alpha \in \left] {0^\circ ,\;90^\circ } \right[\), então \(\cos \alpha > 0\).

Logo, \(\cos \alpha = + \sqrt {\frac{8}{9}} = \frac{{\sqrt 8 }}{3} = \frac{{2\sqrt 2 }}{3}\).

E, finalmente, \({\mathop{\rm tg}\nolimits} \alpha = \frac{{{\mathop{\rm sen}\nolimits} \alpha }}{{\cos \alpha }} = \frac{{\frac{1}{3}}}{{\frac{{2\sqrt 2 }}{3}}} = \frac{1}{3} \times \frac{3}{{2\sqrt 2 }} \times \frac{{\sqrt 2 }}{{\sqrt 2 }} = \frac{{\sqrt 2 }}{4}\).