\(\sin^2\alpha+\cos^2\alpha=1\Leftrightarrow\sin^2\alpha=1-\dfrac{1}{16}=\dfrac{15}{16}\\ \Leftrightarrow\sin\alpha=\dfrac{\sqrt{15}}{4}\\ \cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{1}{4}\cdot\dfrac{4}{\sqrt{15}}=\dfrac{1}{\sqrt{15}}=\dfrac{\sqrt{15}}{15}\)