Answer
$$\frac{{63}}{{65}}$$
Work Step by Step
$$\eqalign{
& \cos \left( {{{\tan }^{ - 1}}\frac{5}{{12}} - {{\tan }^{ - 1}}\frac{3}{4}} \right) \cr
& {\text{Let }}\theta = {\tan ^{ - 1}}\frac{5}{{12}}{\text{ and }}\beta = {\tan ^{ - 1}}\frac{3}{4} \cr
& \cr
& {\text{Therefore,}} \cr
& \theta = {\tan ^{ - 1}}\frac{5}{{12}} \to \tan \theta = \frac{5}{{12}} = \frac{{{\text{opposite side}}}}{{{\text{adjacent side}}}} \cr
& {\text{hypotenuse}} = \sqrt {{5^2} + {{12}^2}} = 13 \cr
& and \cr
& \beta = {\tan ^{ - 1}}\frac{3}{4} \to \tan \theta = \frac{3}{4} = \frac{{{\text{opposite side}}}}{{{\text{adjacent side}}}} \cr
& {\text{hypotenuse}} = \sqrt {{3^2} + {4^2}} = 5 \cr
& \cr
& \cos \left( {{{\tan }^{ - 1}}\frac{5}{{12}} - {{\tan }^{ - 1}}\frac{3}{4}} \right) = \cos \left( {\theta - \beta } \right) \cr
& {\text{Use the cosine of a difference}} \cr
& = \cos \theta \cos \beta + \sin \theta \sin \beta \cr
& = \left( {\frac{{12}}{{13}}} \right)\left( {\frac{4}{5}} \right) + \left( {\frac{5}{{13}}} \right)\left( {\frac{3}{5}} \right) \cr
& {\text{Simplify}} \cr
& = \frac{{63}}{{65}} \cr} $$
