The value of $$\lim_{x \to 0^+} \frac{\cos^{-1}(x - [x]^2) \cdot \sin^{-1}(x - [x]^2)}{x - x^3}$$, where $$[x]$$ denotes the greatest integer $$\leq x$$ is:

We need $$\lim_{x \to 0^+} \frac{\cos^{-1}(x - [x]^2) \cdot \sin^{-1}(x - [x]^2)}{x - x^3}$$, where $$[x]$$ is the greatest integer function.

As $$x \to 0^+$$, we have $$0 < x < 1$$, so $$[x] = 0$$. Therefore $$[x]^2 = 0$$.

The expression $$x - [x]^2 = x - 0 = x$$.

So the limit becomes $$\lim_{x \to 0^+} \frac{\cos^{-1}(x) \cdot \sin^{-1}(x)}{x - x^3}$$.

We can factor the denominator: $$x - x^3 = x(1 - x^2)$$.

So the limit is $$\lim_{x \to 0^+} \frac{\cos^{-1}(x) \cdot \sin^{-1}(x)}{x(1 - x^2)}$$.

We know that as $$x \to 0$$, $$\sin^{-1}(x) \approx x$$, so $$\frac{\sin^{-1}(x)}{x} \to 1$$.

Also as $$x \to 0$$, $$\cos^{-1}(x) \to \cos^{-1}(0) = \frac{\pi}{2}$$, and $$(1 - x^2) \to 1$$.

Therefore the limit equals $$\frac{\pi}{2} \cdot 1 \cdot \frac{1}{1} = \frac{\pi}{2}$$.

This matches Option D: $$\frac{\pi}{2}$$.