1. The problem states that angle $\theta$ is in quadrant IV and $\cos \theta = \frac{4}{5}$. We need to find $\tan \theta$. 2. Recall that in quadrant IV, cosine is positive and sine is negative. 3. Use the Pythagorean identity: $$\sin^2 \theta + \cos^2 \theta = 1$$ 4. Substitute $\cos \theta = \frac{4}{5}$: $$\sin^2 \theta + \left(\frac{4}{5}\right)^2 = 1$$ $$\sin^2 \theta + \frac{16}{25} = 1$$ 5. Solve for $\sin^2 \theta$: $$\sin^2 \theta = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25}$$ 6. Take the square root to find $\sin \theta$: $$\sin \theta = \pm \frac{3}{5}$$ Since $\theta$ is in quadrant IV, $\sin \theta$ is negative: $$\sin \theta = -\frac{3}{5}$$ 7. Calculate $\tan \theta$ using: $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{3}{5}}{\frac{4}{5}} = -\frac{3}{4}$$ Final answer: $\boxed{-\frac{3}{4}}$