1. **Problem Statement:** Find the exact values of the following trigonometric functions: (a) $\tan\left(\frac{\pi}{3}\right)$ (b) $\sin\left(\frac{7\pi}{6}\right)$ (c) $\sec\left(\frac{5\pi}{3}\right)$ 2. **Formulas and Important Rules:** - Tangent is defined as $\tan \theta = \frac{\sin \theta}{\cos \theta}$. - Sine and cosine values on the unit circle correspond to the y- and x-coordinates respectively. - Secant is the reciprocal of cosine: $\sec \theta = \frac{1}{\cos \theta}$. - Reference angles and signs depend on the quadrant. 3. **Calculations:** (a) $\tan\left(\frac{\pi}{3}\right)$: - $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ - $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$ - So, $\tan\left(\frac{\pi}{3}\right) = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$ (b) $\sin\left(\frac{7\pi}{6}\right)$: - $\frac{7\pi}{6}$ is in the third quadrant where sine is negative. - Reference angle is $\frac{7\pi}{6} - \pi = \frac{\pi}{6}$. - $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ - Therefore, $\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}$ (c) $\sec\left(\frac{5\pi}{3}\right)$: - $\frac{5\pi}{3}$ is in the fourth quadrant where cosine is positive. - Reference angle is $2\pi - \frac{5\pi}{3} = \frac{\pi}{3}$. - $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$ - So, $\sec\left(\frac{5\pi}{3}\right) = \frac{1}{\cos\left(\frac{5\pi}{3}\right)} = \frac{1}{\frac{1}{2}} = 2$ 4. **Final Answers:** - (a) $\sqrt{3}$ - (b) $-\frac{1}{2}$ - (c) $2$