1. The problem asks to determine the secant of $\frac{\sqrt{3}}{2}$.\n\n2. The secant function is defined as the reciprocal of the cosine function: $$\sec(\theta) = \frac{1}{\cos(\theta)}.$$\n\n3. We need to find an angle $\theta$ such that $$\cos(\theta) = \frac{\sqrt{3}}{2}.$$\n\n4. From the unit circle, $$\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}.$$\n\n5. Therefore, $$\sec\left(\frac{\pi}{6}\right) = \frac{1}{\cos\left(\frac{\pi}{6}\right)} = \frac{1}{\frac{\sqrt{3}}{2}}.$$\n\n6. Simplify the fraction: $$\sec\left(\frac{\pi}{6}\right) = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{1 \times 2}{\sqrt{3}} = \frac{2}{\sqrt{3}}.$$\n\n7. To rationalize the denominator, multiply numerator and denominator by $\sqrt{3}$: $$\sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}.$$\n\n8. Final answer: $$\boxed{\sec\left(\frac{\pi}{6}\right) = \frac{2\sqrt{3}}{3}}.$$