Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 286: 36

$\frac{3}{4}$

Given: $cot^{-1}x=tan^{-1}\frac{4}{3}$ Consider $tan^{-1}\frac{4}{3}=P$ $tan P=\frac{4}{3}$ Apply trigonometric identity for tangent. $tanx=\frac{opp}{adj}=\frac{4}{3}$ Thus, $hyp=\sqrt {4^{2}+3^{2}}=\sqrt {16+9}=\sqrt {25}=5$ Likewise, $cotP=\frac{adj}{opp}=\frac{3}{4}$ $cot^{-1}x=P$ gives $x=cotP$ Hence, $x=\frac{3}{4}$