Since [tex]\cot^2\theta+1=\csc^2\theta[/tex], you have
[tex]\cot^2\theta=\left(\dfrac{\sqrt5}2\right)^2-1=\dfrac14[/tex]
[tex]\cot\theta=\pm\sqrt{\dfrac14}=\pm\dfrac12[/tex]
Since [tex]\tan\theta>0[/tex], and [tex]\cot\theta=\dfrac1{\tan\theta}[/tex], you know that [tex]\cot\theta[/tex] is also positive, which means
[tex]\cot\theta=\dfrac12[/tex]
