Answer:
[tex]-\frac{3\sqrt{13}}{13}[/tex].
Step-by-step explanation:
Since we are in quadrant two, cosine value is negative while sine value is positive.
We are going to use the Pythagorean Identity: [tex]1+\tan^2(\theta)=\sec^2(\theta)[/tex].
[tex]1+(\frac{-2}{3})^2=\sec^2(\theta)[/tex]
[tex]1+\frac{4}{9}=\sec^2(\theta)[/tex]
[tex]\frac{9+4}{9}=\sec^2(\theta)[/tex]
[tex]\frac{13}{9}=\sec^2(\theta)[/tex]
[tex]\pm \sqrt{\frac{13}{9}}=\sec(\theta)[/tex]
[tex]\pm \frac{\sqrt{13}}{3}=\sec(\theta)[/tex]
Since cosine and secant are reciprocals then they will have the same sign as along as they both exist.
[tex]\sec(\theta)=-\frac{\sqrt{13}}{3}[/tex]
[tex]\cos(\theta)=-\frac{3}{\sqrt{13}}[/tex].
I don't see this answer as I'm going to rationalize the denominator.
[tex]\cos(\theta)=-\frac{3}{\sqrt{13}} \cdot \frac{\sqrt{13}}{\sqrt{13}}[/tex].
[tex]\cos(\theta)=-\frac{3\sqrt{13}}{13}[/tex].
