\(\csc\theta=\dfrac1{\sin\theta}=\dfrac{\sqrt{17}}}4\implies\sin\theta=\dfrac4{\sqrt{17}}\)
Recall that \(\sin^2\theta+\cos^2\theta=1\), which means
\(\cos\theta=\pm\sqrt{1-\sin^2\theta}=\pm\sqrt{1-\dfrac{16}{17}}=\pm\dfrac1{\sqrt{17}}\)
Since \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}>0\), and \(\sin\theta>0\), it follows that \(\cos\theta>0\) too, so you take the positive root:
\(\cos\theta=\dfrac1{\sqrt{17}}\)
Therefore
\(\cot\theta=\dfrac{\cos\theta}{\sin\theta}=\dfrac{\frac1{\sqrt{17}}}{\frac4{\sqrt{17}}}=\dfrac14\)
