91Ó°ÊÓ

The angle \( -\frac{7 \pi}{12} \) can be broken down to \( -\frac{3 \pi}{12}-\frac{4 \pi}{12} \) or \( -\frac{1}{4} \pi-\frac{1}{3} \pi \) which are angles easier to deal with on the unit circle.

Use the formula for the sum of two angles, \( \sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b) \) and \( \cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b) \), to find the sine and cosine. Substituting \( a = -\frac{1}{4} \pi \) and \( b = -\frac{1}{3} \pi \) into these formulas, the sine becomes \( \sin(-\frac{7}{12}\pi) = \sin(-\frac{1}{4}\pi)\cos(-\frac{1}{3}\pi) + \cos(-\frac{1}{4}\pi)\sin(-\frac{1}{3}\pi) = -\frac{\sqrt{2}}{2}*\frac{1}{2} - \frac{\sqrt{2}}{2}*-\frac{\sqrt{3}}{2} = -\frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4} \) and the cosine becomes \( \cos(-\frac{7}{12}\pi) = \cos(-\frac{1}{4}\pi)\cos(-\frac{1}{3}\pi) - \sin(-\frac{1}{4}\pi)\sin(-\frac{1}{3}\pi) = \frac{\sqrt{2}}{2}*\frac{1}{2} - -\frac{\sqrt{2}}{2}*- \frac{\sqrt{3}}{2} = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4} \)

The tangent of the angle can be found by taking the ratio of the sine to the cosine, i.e. \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \). Here, \( \tan(-\frac{7}{12}\pi) = \frac{\sin(-\frac{7}{12}\pi)}{\cos(-\frac{7}{12}\pi)} = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{\frac{\sqrt{2} - \sqrt{6}}{4}} = -1 \)