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