If $$2\sin \left( {\frac{{\pi x}}{2}} \right) = {x^2} + \frac{1}{{{x^2}}}{\text{,}}$$     then the value of $$\left( {x - \frac{1}{x}} \right)$$   is?

$$\eqalign{ & 2\sin \left( {\frac{{\pi x}}{2}} \right) = {x^2} + \frac{1}{{{x^2}}} \cr & {\text{Let }}x = 1 \cr & \Rightarrow 2\sin {90^ \circ } = {1^2} + \frac{1}{{{1^2}}} \cr & \Rightarrow 2 \times 1 = 1 + 1 \cr & 2 = 2\left( {{\text{Matched}}} \right) \cr & {\text{So, }}x = 1 \cr & \Rightarrow \left( {x - \frac{1}{x}} \right) \cr & \Rightarrow 1 - \frac{1}{1} \cr & \Rightarrow 0 \cr} $$