Question

The value of $\frac{1}{\sqrt{12-\sqrt{140}}}-\frac{1}{\sqrt{8-\sqrt{60}}}-\frac{2}{\sqrt{10+\sqrt{84}}}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

Given exp.$=\frac{1}{\sqrt{7+5-\sqrt{4\times 7\times 5}}}-\frac{1}{\sqrt{5+3-\sqrt{4\times 5\times 3}}}$$-\frac{2}{\sqrt{7+3+\sqrt{4\times 7\times 3}}}$

$=\frac{1}{\sqrt{(\sqrt{7}{)}^2+(\sqrt{5}{)}^2-2\sqrt{7\times 5}}}-$$\frac{1}{\sqrt{(\sqrt{5}{)}^2+(\sqrt{3}{)}^2-2\sqrt{5\times 3}}}-$$\frac{2}{\sqrt{(\sqrt{7}{)}^2+(\sqrt{3}{)}^2+2\sqrt{7\times 3}}}$

$=\frac{1}{\sqrt{(\sqrt{7}-\sqrt{5}{)}^2}}-\frac{1}{\sqrt{(\sqrt{5}-\sqrt{3}{)}^2}}-\frac{2}{\sqrt{(\sqrt{7}+\sqrt{3}{)}^2}}$

$=\frac{1}{\sqrt{7}-\sqrt{5}}-\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{2}{\sqrt{7}+\sqrt{3}}$

Rationalising the denominator, we get,

Given exp.$=\frac{\sqrt{7}+\sqrt{5}}{7-5}-\frac{(\sqrt{5}+\sqrt{3})}{5-3}-\frac{2(\sqrt{7}-\sqrt{3})}{7-3}$

$=\frac{\sqrt{7}+\sqrt{5}}{2}-\frac{(\sqrt{5}+\sqrt{3})}{2}-\frac{(\sqrt{7}-\sqrt{3})}{2}$

$\frac{=\sqrt{7}+\sqrt{5}-\sqrt{5}-\sqrt{3}-\sqrt{7}+\sqrt{3}}{2}=\frac{0}{2}=0$