Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 260: 102

$cot~(arcsin~u) = \frac{\sqrt{1-u^2}}{u}$

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