If $$x^2+\frac{1}{x^2}=2$$, then the value of $$x-\frac{1}{x}$$ is
Given : $$x^2+\frac{1}{x^2}=2$$ ----------(i)
Now, $$(x-\frac{1}{x})^2=x^2+\frac{1}{x^2}-2(x)(\frac{1}{x})$$
=> $$(x-\frac{1}{x})^2=2-2$$ Â Â [Using (i)]
=> $$x-\frac{1}{x}=0$$
=> Ans - (B)
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