Question

The value of $\frac{(\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1})(\sqrt{3}-\sqrt{2})}{(\sqrt{\sqrt{3}+1}-\sqrt{\sqrt{3}-1})}$ is

• A. $\frac{(\sqrt{3}+\sqrt{2})}{\sqrt{3}}$
• B. $1$
• C. $\sqrt{3}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is B $1$

$\frac{(\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1})(\sqrt{3}-\sqrt{2})}{(\sqrt{\sqrt{3}+1}-\sqrt{\sqrt{3}-1})}$

Rationalization the above expression,

$=\frac{\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1}}{(\sqrt{\sqrt{3}+1}-\sqrt{\sqrt{3}-1})}\times \frac{\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1}}{\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1}}\times (\sqrt{3}-\sqrt{2})$

$=\frac{(\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1}{)}^2\times (\sqrt{3}-\sqrt{2})}{(\sqrt{\sqrt{3}+1}{)}^2-(\sqrt{\sqrt{3}-1}{)}^2}$ ____ $\because a^2-b^2=(a+b)(a-b)$

$=\frac{(\sqrt{3}+1+\sqrt{3}-1+2(\sqrt{\sqrt{3}+1}\left)\right(\sqrt{\sqrt{3}-1}\left)\right)\times (\sqrt{3}-\sqrt{2})}{(\sqrt{3}+1)-(\sqrt{3}-1)}$

$=\frac{[2\sqrt{3}+2(\sqrt{\sqrt{3}+1}\left)\right(\sqrt{\sqrt{3}-1}\left)\right]\times (\sqrt{3}-\sqrt{2})}{\sqrt{3}+1-\sqrt{3}+1}$

$=\frac{[2\sqrt{3}+2\sqrt{(\sqrt{3}+1)(\sqrt{3}-1)}]\times (\sqrt{3}-\sqrt{2})}{2}$

$=\frac{2(\sqrt{3}+\sqrt{3-1})(\sqrt{3}-\sqrt{2})}{2}$

$=\frac{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}{1}$

$=\frac{3-2}{1}$

$=1$