Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 280: 32

$\frac{12}{5}$

Given: $arctanx=arccos\frac{5}{13}$ Consider $arccos\frac{5}{13}=P$ $cos P=\frac{5}{13}$ Apply trigonometric identity for cosine. $cosx=\frac{adj}{hyp}=\frac{5}{13}$ Thus, $opp=\sqrt {13^{2}-5^{2}}=\sqrt {169-25}=12$ Likewise, $tanP=\frac{opp}{adj}=\frac{12}{5}$ $arctanx=P$ gives $x=tanP$ Hence, $x=\frac{12}{5}$