1. The problem asks to find the value of \( \sec \theta \) when \( 3\pi/2 < \theta < 2\pi \) and \( \sec \theta = \sqrt{5} \). 2. Recall that \( \sec \theta = \frac{1}{\cos \theta} \). 3. Given \( \sec \theta = \sqrt{5} \), we find \( \cos \theta = \frac{1}{\sec \theta} = \frac{1}{\sqrt{5}} \). 4. Since \( \theta \) is in the fourth quadrant \((3\pi/2 < \theta < 2\pi)\), cosine is positive, so \( \cos \theta = \frac{1}{\sqrt{5}} \). 5. Therefore, \( \sec \theta = \sqrt{5} \) as given, confirming the value. Final answer: \( \sec \theta = \sqrt{5} \)