Question

Find the exact value of cos(theta) for an angle (theta) with tan (theta)= -2/3 and with its terminal side in Quadrant II.

Answer 1

Answer:

$-\frac{3\sqrt{13}}{13}$.

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: $1+\tan^2(\theta)=\sec^2(\theta)$.

$1+(\frac{-2}{3})^2=\sec^2(\theta)$

$1+\frac{4}{9}=\sec^2(\theta)$

$\frac{9+4}{9}=\sec^2(\theta)$

$\frac{13}{9}=\sec^2(\theta)$

$\pm \sqrt{\frac{13}{9}}=\sec(\theta)$

$\pm \frac{\sqrt{13}}{3}=\sec(\theta)$

Since cosine and secant are reciprocals then they will have the same sign as along as they both exist.

$\sec(\theta)=-\frac{\sqrt{13}}{3}$

$\cos(\theta)=-\frac{3}{\sqrt{13}}$.

I don't see this answer as I'm going to rationalize the denominator.

$\cos(\theta)=-\frac{3}{\sqrt{13}} \cdot \frac{\sqrt{13}}{\sqrt{13}}$.

$\cos(\theta)=-\frac{3\sqrt{13}}{13}$.

Answer 2

Answer:

b

Step-by-step explanation:

1 + tan²theta = sec²theta

1 + (-2/3)² = sec²theta

1 + 4/9 = sec²theta

13/9 = sec²theta

sec theta = -sqrt(13)/3

Because Quadrant 2

cos theta = 1 ÷ sec theta

Cos there = -3/sqrt(13)

-3sqrt(13)/13

After rationalization