Chapter 7 - Trigonometric Identities and Equations - 7.5 Inverse Circular Functions - 7.5 Exercises - Page 710: 96

$$\frac{{\sqrt {1 - {u^2}} }}{u}$$

$$\eqalign{ & \tan \left( {\arccos u} \right) \cr & {\text{From the triangle we have that}} \cr & \cos \theta = u \cr & \theta = \arccos u \cr & \tan \theta = \left( {\arccos u} \right) = \frac{{{\text{opposite side}}}}{{{\text{adjacent side}}}} \cr & \tan \theta = \frac{{\sqrt {1 - {u^2}} }}{u} \cr} $$