Chapter 4 - Section 4.7 - Inverse Trigonometric Functions - 4.7 Problem Set - Page 263: 88

$\frac{1}{\sqrt{1 + x^2}}$

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