Answer:
[tex]\displaystyle \csc\theta=-\frac{\sqrt{5}}{2}[/tex]
Step-by-step explanation:
Since [tex]\displaystyle \csc\theta=\frac{1}{\sin\theta}=\frac{1}{\frac{\text{Opposite}}{\text{Hypotenuse}}}=\frac{\text{Hypotenuse}}{\text{Opposite}}[/tex], we need to find the hypotenuse given our corresponding opposite and adjacent lengths of -6 and -3 accounting for Quadrant III:
[tex](-6)^2+(-3)^2=c^2\\\\36+9=c^2\\\\45=c^2\\\\\sqrt{45}=c\\\\c=3\sqrt{5}[/tex]
Thus, [tex]\displaystyle \csc\theta=\frac{\text{Hypotenuse}}{\text{Opposite}}=\frac{3\sqrt{5}}{-6}=-\frac{\sqrt{5}}{2}[/tex]