Using inverse functions.
If you know,
[tex]\cos(\theta)=-\dfrac{8}{9}\implies \arccos\Big(-\dfrac{8}{9}\Big)=\theta\approx152.734^{\circ}[/tex]
Knowing [tex]\theta[/tex] we obtain,
[tex]
\csc(\theta)\approx2.183 \\
\cot(\theta)\approx-1.94
[/tex]
Hope this helps.
