Question

Let o be an angle in quadrant I such that sin0=7/10 Find the exact values of sec 0 and cot0.

Answer 1

We will use this rule to solve the question

`\sin ^2O+\cos ^2O=1`

sin O = 7/10, substitute it in the rule to find cos O

`\begin{gathered} ((7)/(10))^2+\cos ^2O=1 \\ (40)/(100)+\cos ^2O=1 \\ \cos ^2O=1-(49)/(100)=(51)/(100) \\ \cos O=\frac{\sqrt[]{51}}{10} \end{gathered}`

sec O = 1/cos O

`\sec O=\frac{10}{\sqrt[]{51}}=\frac{10\sqrt[]{51}}{51}`

cot O = cos O/sin O

`\cot O=(\cos O)/(\sin O)=\frac{\frac{10\sqrt[]{51}}{51}}{(7)/(10)}=2`

cot O = 2