1. The problem is to find the value of $\sec\left(\frac{5\pi}{6}\right)$.\n\n2. Recall that $\sec(\theta) = \frac{1}{\cos(\theta)}$. So we need to find $\cos\left(\frac{5\pi}{6}\right)$ first.\n\n3. The angle $\frac{5\pi}{6}$ radians is in the second quadrant, where cosine values are negative.\n\n4. The reference angle for $\frac{5\pi}{6}$ is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$.\n\n5. We know $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$. Since $\frac{5\pi}{6}$ is in the second quadrant, $\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$.\n\n6. Therefore, $\sec\left(\frac{5\pi}{6}\right) = \frac{1}{\cos\left(\frac{5\pi}{6}\right)} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}}$.\n\n7. To rationalize the denominator, multiply numerator and denominator by $\sqrt{3}$:\n$$-\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}.$$\n\nFinal answer: $$\sec\left(\frac{5\pi}{6}\right) = -\frac{2\sqrt{3}}{3}.$$