Chapter 7 - Trigonometric Identities and Equations - 7.5 Inverse Circular Functions - 7.5 Exercises - Page 710: 104

$$\frac{3}{{\sqrt {9 - {u^2}} }}$$

$$\eqalign{ & \csc \left( {\arctan \frac{{\sqrt {9 - {u^2}} }}{u}} \right) \cr & {\text{From the triangle we have that}} \cr & \tan \theta = \frac{{\sqrt {9 - {u^2}} }}{u} \cr & \theta = \arctan \frac{{\sqrt {9 - {u^2}} }}{u} \cr & \csc \left( {\arctan \frac{{\sqrt {9 - {u^2}} }}{u}} \right) = \csc \theta \cr & \csc \left( {\arctan \frac{{\sqrt {9 - {u^2}} }}{u}} \right) = \frac{{{\text{hypotenuse}}}}{{{\text{opposite side}}}} \cr & \csc \left( {\arctan \frac{{\sqrt {9 - {u^2}} }}{u}} \right) = \frac{3}{{\sqrt {9 - {u^2}} }} \cr} $$