Let $$S = \left\{x \in R : 0 \lt x \lt 1 \text{ and } 2\tan^{-1}\left(\frac{1-x}{1+x}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\}$$. If $$n(S)$$ denotes the number of elements in $$S$$ then :

$$2 \tan^{-1}\left(\frac{1-x}{1+x}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right), \quad 0 < x < 1$$

Using substitution $$x = \tan\theta$$:

$$\text{Since } 0 < x < 1 \implies \theta \in \left(0, \frac{\pi}{4}\right)$$

$$\frac{1-x}{1+x} = \frac{\tan\frac{\pi}{4} - \tan\theta}{1 + \tan\frac{\pi}{4}\tan\theta} = \tan\left(\frac{\pi}{4} - \theta\right)$$

$$\text{LHS} = 2 \tan^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right)$$

Since $$\theta \in \left(0, \frac{\pi}{4}\right) \implies \left(\frac{\pi}{4} - \theta\right) \in \left(0, \frac{\pi}{4}\right)$$:

$$\text{LHS} = 2\left(\frac{\pi}{4} - \theta\right) = \frac{\pi}{2} - 2\theta$$

$$\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right) = \cos^{-1}(\cos 2\theta)$$

Since $$\theta \in \left(0, \frac{\pi}{4}\right) \implies 2\theta \in \left(0, \frac{\pi}{2}\right)$$: $$\text{RHS} = 2\theta$$

$$\frac{\pi}{2} - 2\theta = 2\theta \implies 4\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{8}$$

$$x = \tan\left(\frac{\pi}{8}\right) = \sqrt{2} - 1$$

$$\sqrt{2} - 1 \approx 1.414 - 1 = 0.414 < 0.5$$