Answer:
5/3
Step-by-step explanation:
For this, the useful relation is ...
[tex]\cot^2{(\theta)}+1=\csc^2{(\theta)}\\\\\left(\dfrac{8}{6}\right)^2+1=\csc^2{(\theta)}\\\\\dfrac{64+36}{36}=\csc^2{(\theta)}\\\\\sqrt{\dfrac{100}{36}}=\dfrac{10}{6}=\boxed{\dfrac{5}{3}=\csc{(\theta)}}[/tex]
