1. **State the problem:** We need to find $\tan(\alpha + \beta)$ given $\tan \alpha = \frac{3}{4}$ with $\alpha$ in quadrant I, and $\cos \beta = \frac{4}{5}$ with $\beta$ in quadrant IV. 2. **Recall the formula:** $$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$$ 3. **Find $\tan \beta$:** Since $\cos \beta = \frac{4}{5}$ and $\beta$ is in quadrant IV, $\sin \beta$ is negative. Use Pythagorean identity: $$\sin \beta = -\sqrt{1 - \cos^2 \beta} = -\sqrt{1 - \left(\frac{4}{5}\right)^2} = -\sqrt{1 - \frac{16}{25}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$$ Then, $$\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{-\frac{3}{5}}{\frac{4}{5}} = -\frac{3}{4}$$ 4. **Substitute values into the formula:** $$\tan(\alpha + \beta) = \frac{\frac{3}{4} + \left(-\frac{3}{4}\right)}{1 - \frac{3}{4} \times \left(-\frac{3}{4}\right)} = \frac{0}{1 + \frac{9}{16}} = 0$$ 5. **Final answer:** $$\boxed{0}$$