1. The problem is to find an expression for $\tan 4\theta$ in terms of $\tan \theta$. 2. Use the double-angle formula for tangent: $$\tan 2\alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha}$$ 3. First, express $\tan 2\theta$: $$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$$ 4. Now, apply the double-angle formula again to get $\tan 4\theta$: $$\tan 4\theta = \tan(2 \cdot 2\theta) = \frac{2 \tan 2\theta}{1 - \tan^2 2\theta}$$ 5. Substitute $\tan 2\theta$ from step 3: $$\tan 4\theta = \frac{2 \cdot \frac{2 \tan \theta}{1 - \tan^2 \theta}}{1 - \left(\frac{2 \tan \theta}{1 - \tan^2 \theta}\right)^2}$$ 6. Simplify numerator: $$\frac{4 \tan \theta}{1 - \tan^2 \theta}$$ 7. Simplify denominator: $$1 - \frac{4 \tan^2 \theta}{(1 - \tan^2 \theta)^2} = \frac{(1 - \tan^2 \theta)^2 - 4 \tan^2 \theta}{(1 - \tan^2 \theta)^2}$$ 8. So, $$\tan 4\theta = \frac{\frac{4 \tan \theta}{1 - \tan^2 \theta}}{\frac{(1 - \tan^2 \theta)^2 - 4 \tan^2 \theta}{(1 - \tan^2 \theta)^2}} = \frac{4 \tan \theta (1 - \tan^2 \theta)}{(1 - \tan^2 \theta)^2 - 4 \tan^2 \theta}$$ 9. Expand denominator: $$(1 - \tan^2 \theta)^2 - 4 \tan^2 \theta = 1 - 2 \tan^2 \theta + \tan^4 \theta - 4 \tan^2 \theta = 1 - 6 \tan^2 \theta + \tan^4 \theta$$ 10. Final formula: $$\boxed{\tan 4\theta = \frac{4 \tan \theta (1 - \tan^2 \theta)}{1 - 6 \tan^2 \theta + \tan^4 \theta}}$$ This expresses $\tan 4\theta$ in terms of $\tan \theta$.