If frac{1}{sqrt{a}} - frac{1}{sqrt{b}} = 0,th | vedclass

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(C) Given the equation: $\frac{1}{\sqrt{a}} - \frac{1}{\sqrt{b}} = 0$.This implies $\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{b}}$.Squaring both sides,we ge

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$\frac{2}{\sqrt{ab}}$

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(C) Given the equation: $\frac{1}{\sqrt{a}} - \frac{1}{\sqrt{b}} = 0$.This implies $\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{b}}$.Squaring both sides,we get $\frac{1}{a} = \frac{1}{b}$.Let $\frac{1}{a} = \frac{1}{b} = k$.Then $\frac{1}{a} + \frac{1}{b} = k + k = 2k$.Since $\frac{1}{a} = k$,we have $a = \frac{1}{k}$,and similarly $b = \frac{1}{k}$.Thus,$\sqrt{ab} = \sqrt{\frac{1}{k} \cdot \frac{1}{k}} = \sqrt{\frac{1}{k^2}} = \frac{1}{k}$.Therefore,$k = \frac{1}{\sqrt{ab}}$.Substituting this back into the expression $2k$,we get $2 \cdot \frac{1}{\sqrt{ab}} = \frac{2}{\sqrt{ab}}$.

If $\frac{1}{\sqrt{a}} - \frac{1}{\sqrt{b}} = 0$,then the value of $\frac{1}{a} + \frac{1}{b}$ is

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