[tex]\text{Given that,}~\\\\\cot \theta = \dfrac 1{\sqrt 3} \\\\\\\text{So,}~~ \tan \theta = \dfrac 1{\cot \theta} = \sqrt 3\\\\\\\sec^2 \theta + \csc^2 \theta \\\\=(1+\tan^2 \theta) + (1 + \cot^2 \theta)\\\\=1 +\left (\sqrt 3 \right)^2 + 1 + \left(\dfrac 1{\sqrt 3} \right)^2\\\\=1 +3 + 1 + \dfrac 13 \\\\= 5 + \dfrac 13 \\\\= 5 \dfrac 13[/tex]
