If the sum of all the solutions of $$\tan^{-1}\left(\frac{2x}{1-x^2}\right) + \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{3}$$, $$-1 < x < 1$$, $$x \neq 0$$, is $$\alpha - \frac{4}{\sqrt{3}}$$, then $$\alpha$$ is equal to _____.

Given $$\tan^{-1}\left(\frac{2x}{1-x^2}\right) + \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{3}$$, with $$-1 < x < 1$$, $$x \neq 0$$.

Simplify using standard identities.

For $$|x| < 1$$: $$\tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}(x)$$.

For the $$\cot^{-1}$$ term, we use $$\cot^{-1}(y) = \frac{\pi}{2} - \tan^{-1}(y)$$.

Case 1 — $$0 < x < 1$$.

Here $$\frac{1-x^2}{2x} > 0$$. Note that $$\tan^{-1}\left(\frac{1-x^2}{2x}\right) + \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{2}$$.

So $$\tan^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{2} - 2\tan^{-1}(x)$$.

Therefore $$\cot^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{2} - \left(\frac{\pi}{2} - 2\tan^{-1}(x)\right) = 2\tan^{-1}(x)$$.

The equation becomes:

$$2\tan^{-1}(x) + 2\tan^{-1}(x) = \frac{\pi}{3}$$

$$4\tan^{-1}(x) = \frac{\pi}{3} \implies \tan^{-1}(x) = \frac{\pi}{12}$$

$$x_1 = \tan\frac{\pi}{12} = 2 - \sqrt{3} \approx 0.268$$

Case 2 — $$-1 < x < 0$$.

Here $$\frac{1-x^2}{2x} < 0$$, and we need to be careful with the range of $$\cot^{-1}$$.

Writing $$x = -t$$ with $$t \in (0,1)$$: $$\frac{1-x^2}{2x} = -\frac{1-t^2}{2t}$$.

$$\cot^{-1}\left(-\frac{1-t^2}{2t}\right) = \pi - \cot^{-1}\left(\frac{1-t^2}{2t}\right) = \pi - 2\tan^{-1}(t) = \pi + 2\tan^{-1}(x)$$

The equation becomes:

$$2\tan^{-1}(x) + \pi + 2\tan^{-1}(x) = \frac{\pi}{3}$$

$$4\tan^{-1}(x) = -\frac{2\pi}{3} \implies \tan^{-1}(x) = -\frac{\pi}{6}$$

$$x_2 = -\tan\frac{\pi}{6} = -\frac{1}{\sqrt{3}} \approx -0.577$$

Compute the sum of solutions.

$$x_1 + x_2 = (2 - \sqrt{3}) - \frac{1}{\sqrt{3}} = 2 - \sqrt{3} - \frac{\sqrt{3}}{3} = 2 - \frac{3\sqrt{3} + \sqrt{3}}{3} = 2 - \frac{4\sqrt{3}}{3} = 2 - \frac{4}{\sqrt{3}}$$

Given this equals $$\alpha - \frac{4}{\sqrt{3}}$$, we get $$\alpha = 2$$.

The answer is $$2$$.