1. Problem statement: Find the exact values of (a) $\tan(\frac{\pi}{3})$, (b) $\sin(\frac{7\pi}{6})$, and (c) $\sec(\frac{5\pi}{3})$. 2. For (a) $\tan(\frac{\pi}{3})$: - Recall that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. - Using the unit circle or special angle values, $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$. - Therefore, $\tan(\frac{\pi}{3}) = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$. 3. For (b) $\sin(\frac{7\pi}{6})$: - The angle $\frac{7\pi}{6}$ is in the third quadrant where sine is negative. - Reference angle is $\frac{7\pi}{6} - \pi = \frac{\pi}{6}$. - $\sin(\frac{\pi}{6}) = \frac{1}{2}$, so $\sin(\frac{7\pi}{6}) = -\frac{1}{2}$. 4. For (c) $\sec(\frac{5\pi}{3})$: - Recall $\sec(\theta) = \frac{1}{\cos(\theta)}$. - The angle $\frac{5\pi}{3}$ is in the fourth quadrant where cosine is positive. - Reference angle is $2\pi - \frac{5\pi}{3} = \frac{\pi}{3}$. - $\cos(\frac{\pi}{3}) = \frac{1}{2}$, so $\cos(\frac{5\pi}{3}) = \frac{1}{2}$. - Therefore, $\sec(\frac{5\pi}{3}) = \frac{1}{\frac{1}{2}} = 2$. Final answers: (a) $\tan(\frac{\pi}{3}) = \sqrt{3}$ (b) $\sin(\frac{7\pi}{6}) = -\frac{1}{2}$ (c) $\sec(\frac{5\pi}{3}) = 2$