Chapter 7 - Trigonometric Identities and Equations - 7.5 Inverse Circular Functions - 7.5 Exercises - Page 709: 85

$$2$$

$$\eqalign{ & \sec \left( {{{\sec }^{ - 1}}2} \right) \cr & {\text{Let }}\theta = {\sec ^{ - 1}}2{\text{, thus}} \cr & \sec \theta = 2 \cr & {\text{Then,}} \cr & \sec \left( {{{\sec }^{ - 1}}2} \right) = \sec \theta \cr & \sec \left( {{{\sec }^{ - 1}}2} \right) = 2 \cr} $$