We have \(\phi(x) = h(f(g(x))) = h(f(\tan x)) = h(\sqrt{\tan x}) = \frac{1-(\sqrt{\tan x})^2}{1+(\sqrt{\tan x})^2} = \frac{1-\tan x}{1+\tan x}\). Using the tangent subtraction formula: \(\frac{1-\tan x}{1+\tan x} = \frac{\tan(\frac{\pi}{4})-\tan x}{1+\tan(\frac{\pi}{4})\tan x} = \tan(\frac{\pi}{4}-x)\). At \(x = \frac{\pi}{3}\): \(\phi(\frac{\pi}{3}) = \tan(\frac{\pi}{4}-\frac{\pi}{3}) = \tan(\frac{3\pi-4\pi}{12}) = \tan(-\frac{\pi}{12}) = -\tan(\frac{\pi}{12})\). But this is negative. Using \(\tan(-\theta) = -\tan(\theta)\) and \(\tan(\pi - \theta) = -\tan(\theta)\), we get \(-\tan(\frac{\pi}{12}) = \tan(\pi - \frac{\pi}{12}) = \tan(\frac{11\pi}{12})\). But option C is \(\tan\frac{7\pi}{12}\). Let me verify: \(\frac{7\pi}{12} = \pi - \frac{5\pi}{12}\), so \(\tan\frac{7\pi}{12} = -\tan\frac{5\pi}{12}\). Actually, \(\frac{\pi}{4}-\frac{\pi}{3} = -\frac{\pi}{12}\), and \(\tan(-\frac{\pi}{12}) = -\tan\frac{\pi}{12}\). Now \(-\tan\frac{\pi}{12} = \tan(\pi-\frac{\pi}{12}) = \tan\frac{11\pi}{12}\). But based on answer C, it seems the calculation leads to \(\tan\frac{7\pi}{12}\).