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Given the function $f(x) = \tan x - 4x$. To determine the intervals of increase, we first find the derivative $f'(x)$ with respect to $x$: $$f'(x) = \sec^2 x - 4$$
A function is increasing when $f'(x) > 0$. Therefore, we solve the inequality: $$\sec^2 x - 4 > 0$$ $$\sec^2 x > 4$$
Taking the square root on both sides, and noting that in the interval $(0, \frac{\pi}{2})$, $\sec x$ is positive: $$\sec x > 2$$ $$\frac{1}{\cos x} > 2 \implies \cos x < \frac{1}{2}$$
In the interval $(0, \frac{\pi}{2})$, we know that $\cos x = \frac{1}{2}$ at $x = \frac{\pi}{3}$. Since the cosine function is strictly decreasing in the first quadrant, $\cos x < \frac{1}{2}$ for $x > \frac{\pi}{3}$. Thus, the interval is $(\frac{\pi}{3}, \frac{\pi}{2})$.
Final Answer: (\frac{\pi}{3}, \frac{\pi}{2})