1. The problem asks to evaluate the trigonometric functions for given angles without using a calculator, using the fact that $\pi$ radians equals 180 degrees. 2. Recall the exact values of trigonometric functions for special angles: $\frac{\pi}{6} = 30^\circ$, $\frac{\pi}{4} = 45^\circ$, $\frac{\pi}{3} = 60^\circ$, $\frac{\pi}{2} = 90^\circ$, $\pi = 180^\circ$, $\frac{3\pi}{4} = 135^\circ$. 3. Use the unit circle and definitions: - $\tan \theta = \frac{\sin \theta}{\cos \theta}$ - $\sec \theta = \frac{1}{\cos \theta}$ - $\csc \theta = \frac{1}{\sin \theta}$ - $\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}$ 4. Evaluate each part of question 9: (a) $\tan(\frac{\pi}{6}) = \tan 30^\circ = \frac{1}{\sqrt{3}}$ (b) $\sec \pi = \sec 180^\circ = \frac{1}{\cos 180^\circ} = \frac{1}{-1} = -1$ (c) $\sec(\frac{3\pi}{4}) = \sec 135^\circ = \frac{1}{\cos 135^\circ} = \frac{1}{-\frac{\sqrt{2}}{2}} = -\sqrt{2}$ (d) $\csc(\frac{\pi}{2}) = \csc 90^\circ = \frac{1}{\sin 90^\circ} = 1$ (e) $\cot(\frac{\pi}{4}) = \cot 45^\circ = 1$ (f) $\tan(-\frac{\pi}{4}) = \tan -45^\circ = -1$ 5. Evaluate each part of question 10: (a) $\tan(\frac{\pi}{3}) = \tan 60^\circ = \sqrt{3}$ (b) $\sec(\frac{\pi}{3}) = \sec 60^\circ = \frac{1}{\cos 60^\circ} = \frac{1}{\frac{1}{2}} = 2$ (c) $\cot(\frac{\pi}{3}) = \cot 60^\circ = \frac{1}{\tan 60^\circ} = \frac{1}{\sqrt{3}}$ (d) $\csc(\frac{\pi}{4}) = \csc 45^\circ = \frac{1}{\sin 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}$ (e) $\tan(-\frac{\pi}{6}) = \tan -30^\circ = -\frac{1}{\sqrt{3}}$ (f) $\cos(-\frac{\pi}{3}) = \cos -60^\circ = \cos 60^\circ = \frac{1}{2}$ Final answers: 9(a) $\frac{1}{\sqrt{3}}$, 9(b) $-1$, 9(c) $-\sqrt{2}$, 9(d) $1$, 9(e) $1$, 9(f) $-1$ 10(a) $\sqrt{3}$, 10(b) $2$, 10(c) $\frac{1}{\sqrt{3}}$, 10(d) $\sqrt{2}$, 10(e) $-\frac{1}{\sqrt{3}}$, 10(f) $\frac{1}{2}$