Question

select the correct answer. consider this equation. cos(θ) = 8/9. if θ is an angle in quadrant iv, what is the value of tan(θ)? a. -√17/9 b. √17/8 c. -√17/8 d. √17/9

Explanation:

Step1: Use the Pythagorean identity $\sin^{2}\theta+\cos^{2}\theta = 1$.

We know that $\cos\theta=\frac{8}{9}$, so $\sin^{2}\theta=1 - \cos^{2}\theta=1-(\frac{8}{9})^{2}=1-\frac{64}{81}=\frac{81 - 64}{81}=\frac{17}{81}$, then $\sin\theta=\pm\frac{\sqrt{17}}{9}$.

Step2: Determine the sign of $\sin\theta$ based on the quadrant.

Since $\theta$ is in quadrant IV, $\sin\theta<0$. So $\sin\theta =-\frac{\sqrt{17}}{9}$.

Step3: Use the formula for $\tan\theta$.

The formula for $\tan\theta$ is $\tan\theta=\frac{\sin\theta}{\cos\theta}$. Substitute $\sin\theta =-\frac{\sqrt{17}}{9}$ and $\cos\theta=\frac{8}{9}$ into the formula, we get $\tan\theta=\frac{-\frac{\sqrt{17}}{9}}{\frac{8}{9}}=-\frac{\sqrt{17}}{8}$.

Answer:

C. $-\frac{\sqrt{17}}{8}$