Answer
$$\frac{{48 + 25\sqrt 3 }}{{39}}$$
Work Step by Step
$$\eqalign{
& \tan \left( {{{\cos }^{ - 1}}\frac{{\sqrt 3 }}{2} - {{\sin }^{ - 1}}\left( { - \frac{3}{5}} \right)} \right) \cr
& = \tan \left( {{{\cos }^{ - 1}}\frac{{\sqrt 3 }}{2} + {{\sin }^{ - 1}}\left( {\frac{3}{5}} \right)} \right) \cr
& {\text{Therefore,}} \cr
& \theta = {\cos ^{ - 1}}\frac{{\sqrt 3 }}{2} \to \cos \theta = \frac{{\sqrt 3 }}{2} = \frac{{{\text{adjacent side}}}}{{{\text{hypotenuse}}}} \cr
& {\text{opposite side}} = \sqrt {{2^2} - {{\left( {\sqrt 3 } \right)}^2}} = 1 \cr
& and \cr
& \beta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right) \to \sin \theta = \frac{3}{5} = \frac{{{\text{opposite side}}}}{{{\text{hypotenuse}}}} \cr
& {\text{adjacent side}} = \sqrt {{5^2} - {3^2}} = 4 \cr
& \cr
& \tan \left( {{{\cos }^{ - 1}}\frac{{\sqrt 3 }}{2} + {{\sin }^{ - 1}}\left( {\frac{3}{5}} \right)} \right) = \tan \left( {\theta + \beta } \right) \cr
& {\text{Use the Tangent of a Sum}} \cr
& = \frac{{\tan \theta + \tan \beta }}{{1 - \tan \theta \tan \beta }} \cr
& = \frac{{\frac{1}{{\sqrt 3 }} + \frac{3}{4}}}{{1 - \left( {\frac{1}{{\sqrt 3 }}} \right)\left( {\frac{3}{4}} \right)}} \cr
& {\text{Simplify}} \cr
& = \frac{{48 + 25\sqrt 3 }}{{39}} \cr} $$
