Chapter 4 - Section 4.7 - Inverse Trigonometric Functions - 4.7 Problem Set - Page 262: 76

$ =\frac{4}{3}$

$\tan (\cos ^ {–1} \frac{3}{5})$ let, $\cos ^ {–1} \frac{3}{5} = \theta$ then, $\cos \theta = \frac{3}{5}$ We know, if $\cos \theta = \frac{b}{h}$, then $\tan \theta = \frac{\sqrt{h^2 - b^2}}{b}$ $\tan \theta = \frac{\sqrt{5^2 - 3^2}}{3}$ $ =\frac{4}{3}$