Chapter 7 - Trigonometric Identities and Equations - 7.5 Inverse Circular Functions - 7.5 Exercises - Page 709: 86

$$\sqrt 2 $$

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