[tex]\displaystyle \sin A =\frac{5}{13} \Rightarrow \cos A = \pm \sqrt{1- \sin^2A} = \pm \sqrt{1 -\bigg(\frac{5}{13}\bigg)^2 } = \pm \frac{12}{13} \\\\\\\ \cot A= \frac{\cos A}{\sin A} = \dfrac{\pm \dfrac{12}{13} }{\dfrac{5}{13} } = \pm \frac{12}{5} \\\\\\ \tan A = \frac{1}{\cot A} = \pm \frac{5}{12}[/tex]
