[tex]\csc\theta=\dfrac{\sqrt5}2\implies \sin\theta=\dfrac1{\csc\theta}=\dfrac2{\sqrt5}[/tex]
Since [tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}>0[/tex], and we know that [tex]\sin\theta=\dfrac2{\sqrt5}>0[/tex], we then also know that [tex]\cos\theta>0[/tex].
Recall that
[tex]\sin^2\theta+\cos^2\theta=1\implies\cos\theta=\pm\sqrt{1-\sin^2\theta}[/tex]
We take the positive root because we know that [tex]\cos\theta>0[/tex]. Then
[tex]\cos\theta=\sqrt{1-\left(\dfrac2{\sqrt5}\right)^2}=\sqrt{1-\dfrac45}=\dfrac1{\sqrt5}[/tex]
Now,
[tex]\cot\theta=\dfrac{\cos\theta}{\sin\theta}=\dfrac{\frac1{\sqrt5}}{\frac2{\sqrt5}}=\dfrac12[/tex]
