Answer :
We will use this rule to solve the question
[tex]\sin ^2O+\cos ^2O=1[/tex]sin O = 7/10, substitute it in the rule to find cos O
[tex]\begin{gathered} (\frac{7}{10})^2+\cos ^2O=1 \\ \frac{40}{100}+\cos ^2O=1 \\ \cos ^2O=1-\frac{49}{100}=\frac{51}{100} \\ \cos O=\frac{\sqrt[]{51}}{10} \end{gathered}[/tex]sec O = 1/cos O
[tex]\sec O=\frac{10}{\sqrt[]{51}}=\frac{10\sqrt[]{51}}{51}[/tex]cot O = cos O/sin O
[tex]\cot O=\frac{\cos O}{\sin O}=\frac{\frac{10\sqrt[]{51}}{51}}{\frac{7}{10}}=2[/tex]cot O = 2