[tex]\\ \sf\longmapsto sec^2A-1=tan^2A[/tex]
[tex]\\ \sf\longmapsto tan^2A=\dfrac{(√85)^2}{6^2}-1[/tex]
[tex]\\ \sf\longmapsto tan^2A=\dfrac{85}{36}-1[/tex]
[tex]\\ \sf\longmapsto tan^2A=\dfrac{85-36}{36}[/tex]
[tex]\\ \sf\longmapsto tan^2A=\dfrac{49}{36}[/tex]
[tex]\\ \sf\longmapsto tanA=\dfrac{7}{6}[/tex]
Now
[tex]\\ \sf\longmapsto cotA=\dfrac{1}{tanA}=\dfrac{6}{7}[/tex]
