8. which value of $\theta$ on the unit circle (in radians) has $\tan\theta=-1$?\n$\frac{3pi}{2}$\n$\frac{pi}{…

Explanation:

Step1: Recall tangent formula

$\tan\theta=\frac{\sin\theta}{\cos\theta}$

Step2: Check each option

For $\theta = \frac{3\pi}{2}$, $\tan\frac{3\pi}{2}$ is undefined as $\cos\frac{3\pi}{2}=0$. For $\theta=\frac{\pi}{2}$, $\tan\frac{\pi}{2}$ is undefined as $\cos\frac{\pi}{2}=0$. For $\theta = \frac{3\pi}{4}$, $\sin\frac{3\pi}{4}=\frac{\sqrt{2}}{2}$ and $\cos\frac{3\pi}{4}=-\frac{\sqrt{2}}{2}$, then $\tan\frac{3\pi}{4}=\frac{\sin\frac{3\pi}{4}}{\cos\frac{3\pi}{4}}=\frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}}=- 1$. For $\theta=\pi$, $\tan\pi=\frac{\sin\pi}{\cos\pi}=\frac{0}{-1}=0$. For $\theta=\frac{5\pi}{6}$, $\sin\frac{5\pi}{6}=\frac{1}{2}$ and $\cos\frac{5\pi}{6}=-\frac{\sqrt{3}}{2}$, then $\tan\frac{5\pi}{6}=\frac{\sin\frac{5\pi}{6}}{\cos\frac{5\pi}{6}}=-\frac{\sqrt{3}}{3}$.

Answer:

$\frac{3\pi}{4}$