We know that
[tex]\begin{gathered} \tan \phi=\frac{\sin\phi}{\cos\phi}=\frac{1}{\cot\phi} \\ \cot \phi=\frac{\cos\phi}{\sin\phi}=\frac{1}{\tan\phi} \\ \sin ^2\phi+\cos ^2\phi=1 \end{gathered}[/tex]Using the third equation we are going to find cos∅:
[tex]\begin{gathered} \sin ^2\phi+\cos ^2\phi=1 \\ \cos ^2\phi=1-\sin ^2\phi \\ \cos ^2\phi=1-(-\frac{4}{5})^2 \\ \cos ^2\phi=1-\frac{16}{25} \\ \cos ^2\phi=\frac{9}{25} \\ \cos \phi=\pm\sqrt[]{\frac{9}{25}} \\ \cos \phi=\pm\frac{3}{5} \end{gathered}[/tex]We, have two possibilities cos∅ is positive or cos∅ is negative.
By the first equation and the given information Tan∅ > 0, we know that
[tex]\frac{\sin \phi}{\cos \phi}>0[/tex]Since sin∅ is negative, and sin∅/cos∅ is positive, cos∅ must be negative so tan∅ > 0. Then
[tex]\cos \phi=-\frac{3}{5}[/tex]Using the second equation, we have that:
[tex]\begin{gathered} \cot \phi=\frac{\cos \phi}{\sin \phi} \\ =\frac{-\frac{3}{5}}{-\frac{4}{5}}=\frac{3}{5}\cdot\frac{5}{4} \\ =\frac{3}{4} \end{gathered}[/tex]