If (2+sqrt{3}) a=(2-sqrt{3}) b=1, then the va | vedclass

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(D) Given that $(2+\sqrt{3}) a = 1$ and $(2-\sqrt{3}) b = 1$.From $(2+\sqrt{3}) a = 1$,we get $a = \frac{1}{2+\sqrt{3}}$.Therefore,$\frac{1}{a} = 2+\s

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$4$

Vedclass · Answer

(D) Given that $(2+\sqrt{3}) a = 1$ and $(2-\sqrt{3}) b = 1$.From $(2+\sqrt{3}) a = 1$,we get $a = \frac{1}{2+\sqrt{3}}$.Therefore,$\frac{1}{a} = 2+\sqrt{3}$.Similarly,from $(2-\sqrt{3}) b = 1$,we get $b = \frac{1}{2-\sqrt{3}}$.Therefore,$\frac{1}{b} = 2-\sqrt{3}$.Now,adding the two values:$\frac{1}{a} + \frac{1}{b} = (2+\sqrt{3}) + (2-\sqrt{3}) = 4$.

If $(2+\sqrt{3}) a=(2-\sqrt{3}) b=1,$ then the value of $\frac{1}{a}+\frac{1}{b}$ is

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