Answer:
[tex]\large\boxed{-\dfrac{8}{\sqrt{63}}}[/tex]
Step-by-step explanation:
[tex]\csc\theta=\dfrac{1}{\sin\theta}\\\\\cos\theta=-\dfrac{1}{8}\qquad\text{use}\ \sin^2\theta+\cos^2\theta=1\\\\\sin^2\theta+\left(-\dfrac{1}{8}\right)^2=1\\\\\sin^2\theta+\dfrac{1}{64}=1\qquad\text{subtract}\ \dfrac{1}{64}\ \text{from both sides}\\\\\sin^2\theta=\dfrac{64}{64}-\dfrac{1}{64}\\\\\sin^2\theta=\dfrac{63}{64}\to\sin\theta=\pm\sqrt{\dfrac{63}{64}}\\\\180^o<\theta<270^o\to III\ quadrant,\ \sin\theta<0.\\\\\text{Therefore}\ \sin\theta=-\sqrt{\dfrac{63}{64}}=-\dfrac{\sqrt{63}}{\sqrt{64}}=-\dfrac{\sqrt{63}}{8}.[/tex]
[tex]\csc\theta=\dfrac{1}{-\frac{\sqrt{63}}{8}}=-\dfrac{8}{\sqrt{63}}[/tex]
