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

Question

Dr. Abubacker · Accepted Answer

(a)
Given, $x = \frac{1}{2}\left( \sqrt{3} + \frac{1}{\sqrt{3}} ight)$
$	herefore x^{2} = \frac{1}{4}\left( 3 + \frac{1}{3} + 2 ight) = \frac{4}{3}$
Now, $\frac{\sqrt{x^{2} - 1}}{x - \sqrt{x^{2} - 1}} 	imes \frac{x + \sqrt{x^{2} - 1}}{x + \sqrt{x^{2} - 1}}$
$= \frac{x\sqrt{x^{2} - 1} + (x^{2} - 1)}{1}$
$= \frac{1}{2}\left( \sqrt{3} + \frac{1}{\sqrt{3}} ight)\sqrt{\frac{4}{3} - 1} + \left( \frac{4}{3} - 1 ight)$
$= \frac{1}{2}\left( \frac{4}{\sqrt{3}} ight)\frac{1}{\sqrt{3}} + \frac{1}{3} = \frac{2}{3} + \frac{1}{3} = 1$

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

Explanation

Reviewed by Dr. Abubacker

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