Step-by-step Solution:

\[ f'(x) = \frac{1(1 + x \tan x) - x(x \sec^2 x + \tan x)}{(1 + x \tan x)^2} \] \[ = \frac{1 - x^2 \sec^2 x}{(1 + x \tan x)^2} \] Here we see that when \(x^2 \sec^2 x < 1\), \(f'(x) > 0\), and when \(x^2 \sec^2 x > 1\), \(f'(x) < 0\). Hence, the function is maximum at \(x \sec x = 1\). \[ \Rightarrow \cos x = x \]