Chapter 4 - Section 4.7 - Inverse Trigonometric Functions - 4.7 Problem Set - Page 262: 86

$\frac{\sqrt{1 – x^2}}{x}$

Let $\cos ^ {–1} x = \theta$ then, $\cos \theta = x$ To evaluate $\tan (\cos ^ {–1} x) = \tan \theta$ We know that, $\tan \theta = \sqrt{\sec ^2 \theta - 1}$ also $\sec \theta = \frac{1}{\cos \theta} $ Using above relations we get $\tan \theta = \sqrt{\frac{1}{x^2} - 1}$ $\tan \theta = \frac{\sqrt{1 – x^2}}{x}$