Answer
See below
Work Step by Step
$\cot \theta=\dfrac{1}{\dfrac{7}{3}}=\dfrac{3}{7}$
$\tan \theta=\dfrac{\sin \theta}{\cos \theta}\\\rightarrow\sin \theta=\tan \theta \times\cos \theta\\\rightarrow \sin \theta=\dfrac{7}{3}\cos \theta$
From the Pythagorean identity, we obtain:
$$\sin^2 \theta +\cos^2 \theta=1\\(\dfrac{7}{3}\cos \theta)^2+\cos^2 \theta=1\\\dfrac{49}{9}\cos^2+\cos^2 \theta=1\\ \dfrac{58}{9}\cos^2 \theta=1\\\cos^2\theta=\frac{9}{58}\\\\\cos \theta=\sqrt \frac{9}{58}=\frac{3\sqrt 58}{58}$$
$\sin \theta=\dfrac{7}{3}\times\dfrac{3\sqrt 58}{58}=\dfrac{7\sqrt 58}{58}$
$\sec \theta=\dfrac{1}{\dfrac{3\sqrt 58}{58}}=\dfrac{\sqrt 58}{3}$
$\csc \theta=\dfrac{1}{\dfrac{7\sqrt 58}{58}}=\dfrac{58}{7\sqrt 58}=\dfrac{\sqrt 58}{7}$
