Answer
See below
Work Step by Step
$\csc \theta=\dfrac{1}{\dfrac{5}{6}}=\dfrac{6}{5}$
From the Pythagorean identity, we obtain:
$\sin^2 \theta +\cos^2 \theta=1\\\cos^2\theta=1-\sin^2 \theta\\\cos \theta = \sqrt 1-\sin^2\theta=\sqrt 1 -(\dfrac{5}{6})^2=\sqrt \frac{11}{36}=\dfrac{\sqrt 11}{6}$
$\sec \theta=\dfrac{1}{\frac{\sqrt 11}{6}}=\dfrac{6}{\sqrt 11}=\dfrac{6\sqrt 11}{11}$
$\tan \theta=\dfrac{\frac{5}{6}}{\frac{\sqrt 11}{6}}=\dfrac{5}{\sqrt 11}=\dfrac{5\sqrt 11}{11}$
$\cot \theta=\dfrac{1}{\dfrac{11}{5\sqrt 11}}=\dfrac{11}{5\sqrt 11}=\dfrac{\sqrt 11}{5}$
