If x = frac{1}{2} left( sqrt{3} + frac{1}{sqr | vedclass

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(A) Given $x = \frac{1}{2} \left( \sqrt{3} + \frac{1}{\sqrt{3}} \right) = \frac{1}{2} \left( \frac{3+1}{\sqrt{3}} \right) = \frac{4}{2\sqrt{3}} = \fra

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(A) Given $x = \frac{1}{2} \left( \sqrt{3} + \frac{1}{\sqrt{3}} \right) = \frac{1}{2} \left( \frac{3+1}{\sqrt{3}} \right) = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}}$.Then $x^2 = \left( \frac{2}{\sqrt{3}} \right)^2 = \frac{4}{3}$.Now,$x^2 - 1 = \frac{4}{3} - 1 = \frac{1}{3}$.So,$\sqrt{x^2 - 1} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$.Substituting these values into the expression $\frac{\sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}}$:$= \frac{\frac{1}{\sqrt{3}}}{\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}}} = \frac{\frac{1}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} = 1$.

If $x = \frac{1}{2} \left( \sqrt{3} + \frac{1}{\sqrt{3}} \right)$,then the value of $\frac{\sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}}$ is equal to

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