[tex] sin^{2}_a+cos^{2}_a=1 \Rightarrow cos_a=\sqrt{1-(\dfrac{1}{3})^{2}} = \sqrt{1-\dfrac{1}{9}}=\sqrt{\dfrac{8}{9}}=\dfrac{2 \sqrt{2}}{3} \\ \\ ctg_a=\dfrac{cos_a}{sin_a}= \dfrac{ \dfrac{2 \sqrt{2}}{3}}{ \dfrac{1}{3}} = \dfrac{2 \sqrt{2}}{3} \cdot \dfrac{3}{1} = 2 \sqrt{2} [/tex]
