Ta có: ${\sin}^2x+{\cos }^2x=1$
$\sin x=\dfrac{1}{5}$
$\Rightarrow {\cos}^2x=1-{\sin}^2x$
$=1-\dfrac{1}{25}=\dfrac{24}{5}$
$\Rightarrow \cos x=\pm\dfrac{2\sqrt6}{5}$
$\Rightarrow \tan x=\dfrac{\sin x}{\cos x}=\dfrac{\pm1}{2\sqrt6}$
$\Rightarrow \cot x=\dfrac{1}{\tan x}=\pm2\sqrt6$