The maximum value of $$y = \tan^{-1}\frac{1-x}{1+x}$$ on [0, 1] is:

Correct option is A

​$$\text{We know from trigonometry that:} \\\frac{1 - x}{1 + x} \text{ is related to the tangent of a difference of angles. Specifically, for } x = \tan(\theta), \text{ the following identity holds:} \\\tan\left( \frac{\pi}{4} - \theta \right) = \frac{1 - \tan(\theta)}{1 + \tan(\theta)} \\\text{Thus, we can rewrite the given expression as:} \\y = \tan^{-1}\left( \frac{1 - x}{1 + x} \right) = \frac{\pi}{4} - \tan^{-1}(x) \\\textbf{Analyze on the interval } x \in [0, 1] \\\text{Let's analyze the behavior of } y \text{ on the interval } x \in [0, 1]. \\\text{When } x = 0: \\y = \frac{\pi}{4} - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \\\text{When } x = 1: \\y = \frac{\pi}{4} - \tan^{-1}(1) = \frac{\pi}{4} - \frac{\pi}{4} = 0$$

Maximum value - π/4