Question

Hi, can you help me to solve this exercise, please!!

Answer 1

The tangent is equal to the sine over the cosine, and the secant is equal to the inverse of the cosine.

Since the angle is in Quadrant II, the sine is positive, the cosine is negative and the secant is negative.

In order to calculate the secant, we can first calculate the cosine using the following equations:

`\begin{gathered} \tan (\theta)=(\sin(\theta))/(\cos(\theta)) \\ -\frac{2\sqrt[]{3}}{3}=(\sin(\theta))/(\cos(\theta)) \\ \sin (\theta)=-\frac{2\sqrt[]{3}}{3}\cos (\theta) \\ \\ \sin ^2(\theta)+\cos ^2(\theta)=1 \\ (-\frac{2\sqrt[]{3}}{3}\cos (\theta))^2+\cos ^2(\theta)=1 \\ (4)/(3)\cos ^2(\theta)+\cos ^2(\theta)=1 \\ (7)/(3)\cos ^2(\theta)=1 \\ \cos ^2(\theta)=(3)/(7) \\ \cos (\theta)=-\frac{\sqrt[]{3}}{\sqrt[]{7}} \\ \\ \sec (\theta)=(1)/(\cos(\theta))=-\frac{\sqrt[]{7}}{\sqrt[]{3}}=-\frac{\sqrt[]{21}}{3} \end{gathered}`

Therefore the value of sec(theta) is equal to -(√21)/3.