$${(\frac{x+1}{x-1})}^{2}$$If $${x}\neq{0}$$ and $${x}\neq{1}$$, and if x is replaced by $$\frac{1}{x}$$ everywhere in the expression above, then the resulting expression is equivalent to
-
A$$(\frac{x+1}{x-1})^{2}$$
-
B$$(\frac{x-1}{x-1})^{2}$$
-
C$$\frac{x^2-1}{1-x^2}$$
-
D$$\frac{x^2-1}{x^2+1}$$
-
E$$-(\frac{x-1}{x+1})^{2}$$