Answer
$tan(arcsin\frac{3}{5}+arccos\frac{5}{7}) = \frac{588+125\sqrt{24}}{184}$
Work Step by Step
Let $A = arcsin\frac{3}{5}$
Then:
$tan~A = \frac{3}{\sqrt{5^2-3^2}} = \frac{3}{4}$
Let $B = arccos(\frac{5}{7})$
Then:
$tan~B = \frac{\sqrt{7^2-5^2}}{5} = \frac{\sqrt{24}}{5}$
We can find $~~tan(arcsin\frac{3}{5}+arccos\frac{5}{7})$:
$tan(A+B) = \frac{tan~A+tan~B}{1-tan~A~tan~B}$
$tan(A+B) = \frac{\frac{3}{4}+\frac{\sqrt{24}}{5}}{1-(\frac{3}{4})(\frac{\sqrt{24}}{5})}$
$tan(A+B) = \frac{\frac{15+4\sqrt{24}}{20}}{\frac{20}{20}-\frac{3\sqrt{24}}{20}}$
$tan(A+B) = (\frac{15+4\sqrt{24}}{20-3\sqrt{24}})~(\frac{20+3\sqrt{24}}{20+3\sqrt{24}})$
$tan(A+B) = \frac{588+125\sqrt{24}}{184}$
Therefore, $~~tan(arcsin\frac{3}{5}+arccos\frac{5}{7}) = \frac{588+125\sqrt{24}}{184}$
