Answer
See below
Work Step by Step
$\tan \theta =\dfrac{1}{\dfrac{6}{11}}=\dfrac{11}{6}$
$\tan \theta=\frac{\sin \theta}{\cos \theta}\\\rightarrow \sin \theta=\dfrac{11}{6}\cos \theta$
From the Pythagorean identity, we obtain:
$$\sin^2 \theta +\cos^2 \theta=1\\(\dfrac{11}{6}\cos \theta)^2+\cos^2 \theta=1\\\cos^2 \theta=1-\dfrac{157}{36}\\\cos \theta=\sqrt\frac{36}{157}\\\\\cos \theta=\frac{6\sqrt 157}{157}$$
$\sec \theta=\dfrac{1}{\dfrac{6\sqrt 157}{157}}=\dfrac{\sqrt 157}{6}$
$\sin \theta=\dfrac{11}{6}\times\dfrac{6\sqrt 157}{157}=\dfrac{11\sqrt 157}{157}$
$\csc \theta=\dfrac{1}{\dfrac{11\sqrt 157}{157}}=\dfrac{\sqrt 157}{11}$
