6. \( \tan \left(-\frac{7 \pi}{12}\right)= \)

To find \( \tan \left(-\frac{7 \pi}{12}\right) \), we can utilize the property of the tangent function that states \( \tan(-x) = -\tan(x) \). Thus, we can simplify: \[ \tan \left(-\frac{7 \pi}{12}\right) = -\tan \left(\frac{7 \pi}{12}\right) \] Next, we can express \( \frac{7 \pi}{12} \) as: \[ \frac{7 \pi}{12} = \frac{3 \pi}{4} - \frac{\pi}{3} \] Using the tangent subtraction formula, \( \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \), we set \( a = \frac{3 \pi}{4} \) and \( b = \frac{\pi}{3} \). We find: \[ \tan \left(\frac{3 \pi}{4}\right) = -1 \quad \text{and} \quad \tan \left(\frac{\pi}{3}\right) = \sqrt{3} \] Now we can substitute these values into the tangent subtraction formula: \[ \tan \left(\frac{7 \pi}{12}\right) = \frac{-1 - \sqrt{3}}{1 + (-1)(\sqrt{3})} = \frac{-1 - \sqrt{3}}{1 - \sqrt{3}} \] To simplify, we can multiply the numerator and denominator by the conjugate of the denominator: \[ \tan \left(\frac{7 \pi}{12}\right) = \frac{(-1 - \sqrt{3})(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{-1 - \sqrt{3} + \sqrt{3} + 3}{1 - 3} = \frac{2}{-2} = -1 \] Thus, \[ \tan \left(\frac{7 \pi}{12}\right) = -1 \] Then, putting that back into our earlier result: \[ \tan \left(-\frac{7 \pi}{12}\right) = -\tan \left(\frac{7 \pi}{12}\right) = -(-1) = 1 \] So, the final answer is: \[ \boxed{1} \]