Question

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

• A. 1
• B. 2
• C. 3
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

Given that
$x=\frac{1}{2}\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)$
or $x^2=\frac{1}{4}\left(3+\frac{1}{3}+2\right)=\frac{4}{3}$
Now, $\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}}=\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\times \frac{x+\sqrt{x^2-1}}{x+\sqrt{x^2-1}}$
$=\frac{x\sqrt{x^2-1}+\left(x^2-1\right)}{1}$
$=\frac{1}{2}\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)\sqrt{\frac{4}{3}-1}+\left(\frac{4}{3}-1\right)$
$=\frac{1}{2}\left(\frac{4}{\sqrt{3}}\right)\frac{1}{\sqrt{3}}+\frac{1}{3}$
$\Rightarrow \frac{2}{3}+\frac{1}{3}=1$