Recall that [tex]\cos^2\theta+\sin^2\theta=1[/tex]. So
[tex]\sin\theta=\pm\sqrt{1-\cos^2\theta}[/tex]
Given that both [tex]\cos\theta<0[/tex] and [tex]\tan\theta<0[/tex], and knowing that [tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}[/tex], it follows that we should expect [tex]\sin\theta>0[/tex], so we take the positive root above.
Now
[tex]\sin\theta=\sqrt{1-\left(-\dfrac15\right)^2}=\dfrac{2\sqrt6}5[/tex]
Then
[tex]\cot\theta=\dfrac{\cos\theta}{\sin\theta}=\dfrac{-\frac15}{\frac{2\sqrt6}5}=-\dfrac1{2\sqrt6}[/tex]
