1. **State the problem:** We need to find the value of $\tan(\arccos(\frac{5}{13}) + \arcsin(\frac{3}{5}))$. 2. **Recall the formula for tangent of a sum:** $$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$ where $A = \arccos(\frac{5}{13})$ and $B = \arcsin(\frac{3}{5})$. 3. **Find $\tan A$:** Since $A = \arccos(\frac{5}{13})$, $\cos A = \frac{5}{13}$. Using Pythagorean identity: $$\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$$ Therefore, $$\tan A = \frac{\sin A}{\cos A} = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{5}$$ 4. **Find $\tan B$:** Since $B = \arcsin(\frac{3}{5})$, $\sin B = \frac{3}{5}$. Using Pythagorean identity: $$\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$ Therefore, $$\tan B = \frac{\sin B}{\cos B} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$$ 5. **Apply the tangent sum formula:** $$\tan(A + B) = \frac{\frac{12}{5} + \frac{3}{4}}{1 - \frac{12}{5} \times \frac{3}{4}} = \frac{\frac{48}{20} + \frac{15}{20}}{1 - \frac{36}{20}} = \frac{\frac{63}{20}}{1 - \frac{36}{20}} = \frac{\frac{63}{20}}{\frac{20}{20} - \frac{36}{20}} = \frac{\frac{63}{20}}{-\frac{16}{20}} = \frac{63}{20} \times \left(-\frac{20}{16}\right) = -\frac{63}{16}$$ 6. **Final answer:** $$\boxed{-\frac{63}{16}}$$