The sum of all the elements in the range of $$f(x) =Sgn(\sin x) + Sgn(\cos x) +Sgn(\tan x) +Sg n(\cot x)$$, $$x \neq \frac{n\pi}{2}, n\epsilon Z, \text{ where } Sgn(t)=\begin{cases}1, & \text{ if } t>0\\-1 & \text{ if } t<0\end{cases} ,is:$$

We need to find the sum of all elements in the range of: $$f(x) = \text{Sgn}(\sin x) + \text{Sgn}(\cos x) + \text{Sgn}(\tan x) + \text{Sgn}(\cot x)$$ where $$x \neq \frac{n\pi}{2}$$ and $$\text{Sgn}(t) = \begin{cases} 1, & t > 0 \\ -1, & t < 0 \end{cases}$$.

Noting that $$\text{Sgn}(\tan x) = \text{Sgn}(\sin x) \cdot \text{Sgn}(\cos x)$$ and that $$\text{Sgn}(\cot x) = \text{Sgn}(\cos x) \cdot \text{Sgn}(\sin x)$$, it follows that $$\text{Sgn}(\tan x) = \text{Sgn}(\cot x)$$.

Introducing the variables $$a = \text{Sgn}(\sin x) = \pm 1$$ and $$b = \text{Sgn}(\cos x) = \pm 1$$ allows us to express the function as $$f(x) = a + b + 2ab$$.

In Quadrant I where $$a = 1$$ and $$b = 1$$, $$f = 4$$; in Quadrant II where $$a = 1$$ and $$b = -1$$, $$f = -2$$; in Quadrant III where $$a = -1$$ and $$b = -1$$, $$f = 0$$; and in Quadrant IV where $$a = -1$$ and $$b = 1$$, $$f = -2$$.

Thus the range of $$f$$ is $$\{-2,0,4\}$$ and the sum of these values is $$-2 + 0 + 4 = 2$$.

The answer is Option D: $$2$$.