Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 266: 96

$tan~(arccos~u) = \frac{\sqrt{1-u^2}}{u}$

Let $~~\theta = arccos~u$ Then $~~cos~\theta = u$ $tan~\theta = \frac{sin~\theta}{cos~\theta}$ $tan~\theta = \frac{\sqrt{1-cos^2~\theta}}{cos~\theta}$ $tan~\theta = \frac{\sqrt{1-u^2}}{u}$ Therefore, $~~tan~(arccos~u) = \frac{\sqrt{1-u^2}}{u}$