1. **Enunciado do problema:** Mostrar que $$f\left(\frac{5\pi}{12}\right) + f\left(-\frac{\pi}{3}\right) = \frac{\sqrt{3}}{3}$$ para a função $$f(x) = \frac{1}{2} \tan\left(\frac{\pi}{6} - 2x\right)$$. 2. **Substituir os valores na função:** Calcule $$f\left(\frac{5\pi}{12}\right)$$: $$f\left(\frac{5\pi}{12}\right) = \frac{1}{2} \tan\left(\frac{\pi}{6} - 2 \times \frac{5\pi}{12}\right) = \frac{1}{2} \tan\left(\frac{\pi}{6} - \frac{5\pi}{6}\right) = \frac{1}{2} \tan\left(-\frac{4\pi}{6}\right) = \frac{1}{2} \tan\left(-\frac{2\pi}{3}\right)$$ Calcule $$f\left(-\frac{\pi}{3}\right)$$: $$f\left(-\frac{\pi}{3}\right) = \frac{1}{2} \tan\left(\frac{\pi}{6} - 2 \times \left(-\frac{\pi}{3}\right)\right) = \frac{1}{2} \tan\left(\frac{\pi}{6} + \frac{2\pi}{3}\right) = \frac{1}{2} \tan\left(\frac{\pi}{6} + \frac{4\pi}{6}\right) = \frac{1}{2} \tan\left(\frac{5\pi}{6}\right)$$ 3. **Avaliar as tangentes:** Sabemos que: $$\tan\left(-\frac{2\pi}{3}\right) = -\tan\left(\frac{2\pi}{3}\right) = -\left(-\sqrt{3}\right) = \sqrt{3}$$ $$\tan\left(\frac{5\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}$$ 4. **Substituir os valores:** $$f\left(\frac{5\pi}{12}\right) = \frac{1}{2} \times \sqrt{3} = \frac{\sqrt{3}}{2}$$ $$f\left(-\frac{\pi}{3}\right) = \frac{1}{2} \times \left(-\frac{1}{\sqrt{3}}\right) = -\frac{1}{2\sqrt{3}}$$ 5. **Somar os resultados:** $$f\left(\frac{5\pi}{12}\right) + f\left(-\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} - \frac{1}{2\sqrt{3}} = \frac{3}{2\sqrt{3}} - \frac{1}{2\sqrt{3}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$ **Resposta final:** $$f\left(\frac{5\pi}{12}\right) + f\left(-\frac{\pi}{3}\right) = \frac{\sqrt{3}}{3}$$