1. **State the problem:** Find the exact value of $\tan\left(-\frac{5\pi}{12}\right)$ using compound angle formulas. 2. **Recall the formula:** The tangent of a sum or difference of angles is given by $$\tan(a \pm b) = \frac{\tan a \pm \tan b}{1 \mp \tan a \tan b}$$ 3. **Choose angles:** Express $-\frac{5\pi}{12}$ as a difference of angles whose tangents are known. For example, $$-\frac{5\pi}{12} = -\left(\frac{\pi}{3} + \frac{\pi}{4}\right) = -\frac{\pi}{3} - \frac{\pi}{4}$$ 4. **Use tangent of sum:** Since $\tan(-x) = -\tan x$, we have $$\tan\left(-\frac{5\pi}{12}\right) = -\tan\left(\frac{\pi}{3} + \frac{\pi}{4}\right)$$ 5. **Calculate $\tan\left(\frac{\pi}{3} + \frac{\pi}{4}\right)$:** $$\tan\left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \frac{\tan \frac{\pi}{3} + \tan \frac{\pi}{4}}{1 - \tan \frac{\pi}{3} \tan \frac{\pi}{4}}$$ 6. **Substitute known values:** $$\tan \frac{\pi}{3} = \sqrt{3}, \quad \tan \frac{\pi}{4} = 1$$ 7. **Evaluate numerator and denominator:** $$\text{numerator} = \sqrt{3} + 1$$ $$\text{denominator} = 1 - \sqrt{3} \times 1 = 1 - \sqrt{3}$$ 8. **Write the fraction:** $$\tan\left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \frac{\sqrt{3} + 1}{1 - \sqrt{3}}$$ 9. **Rationalize the denominator:** Multiply numerator and denominator by the conjugate $1 + \sqrt{3}$: $$\frac{\sqrt{3} + 1}{1 - \sqrt{3}} \times \frac{1 + \sqrt{3}}{1 + \sqrt{3}} = \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{(1)^2 - (\sqrt{3})^2}$$ 10. **Calculate numerator:** $$(\sqrt{3} + 1)(1 + \sqrt{3}) = \sqrt{3} \times 1 + \sqrt{3} \times \sqrt{3} + 1 \times 1 + 1 \times \sqrt{3} = \sqrt{3} + 3 + 1 + \sqrt{3} = 4 + 2\sqrt{3}$$ 11. **Calculate denominator:** $$1 - 3 = -2$$ 12. **Write the fraction:** $$\frac{4 + 2\sqrt{3}}{-2} = \frac{\cancel{2}(2 + \sqrt{3})}{\cancel{2}(-1)} = - (2 + \sqrt{3})$$ 13. **Recall step 4:** $$\tan\left(-\frac{5\pi}{12}\right) = - \tan\left(\frac{\pi}{3} + \frac{\pi}{4}\right) = - \left[-(2 + \sqrt{3})\right] = 2 + \sqrt{3}$$ **Final answer:** $$\boxed{2 + \sqrt{3}}$$