1. Problem: If angle $\theta$ is in quadrant IV and $\cos \theta = \frac{4}{5}$, find $\tan \theta$. 2. Since $\cos \theta = \frac{4}{5}$ and $\theta$ is in quadrant IV, $\sin \theta$ is negative. 3. Use the Pythagorean identity: $$\sin^2 \theta + \cos^2 \theta = 1$$ 4. Substitute $\cos \theta$: $$\sin^2 \theta + \left(\frac{4}{5}\right)^2 = 1$$ 5. Simplify: $$\sin^2 \theta + \frac{16}{25} = 1 \implies \sin^2 \theta = 1 - \frac{16}{25} = \frac{9}{25}$$ 6. Since $\theta$ is in quadrant IV, $\sin \theta = -\frac{3}{5}$. 7. Calculate $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{3}{5}}{\frac{4}{5}} = -\frac{3}{4}$. Final answer: $\boxed{-\frac{3}{4}}$