Answer
See below
Work Step by Step
$\sec \theta=\dfrac{1}{\dfrac{5}{8}}=\dfrac{8}{5}$
From the Pythagorean identity, we obtain:
$\sin^2 \theta +\cos^2 \theta=1\\\cos^2\theta=1-\sin^2 \theta\\\cos \theta = \sqrt 1-\sin^2\theta=\sqrt 1 -(\dfrac{5}{8})^2=\sqrt \frac{39}{64}=\dfrac{\sqrt 39}{8}$
$\csc \theta=\dfrac{1}{\dfrac{\sqrt 39}{8}}=\dfrac{8}{\sqrt 39}=\dfrac{8\sqrt 39}{39}$
$\tan \theta=\dfrac{\dfrac{\sqrt 39}{8}}{\dfrac{5}{8}}=\dfrac{\sqrt 39}{5}$
$\cot \theta=\dfrac{1}{\dfrac{\sqrt 39}{5}}=\dfrac{5}{\sqrt 39}=\dfrac{5\sqrt 39}{39}$
