Question

Evaluate $\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}}$, when $2x=\sqrt{a}+\frac{1}{\sqrt{a}}$

Answer 1

Given, $2x=\sqrt{a}+\frac{1}{\sqrt{a}}$

On squaring and subtracting $4$ on both sides, we get

$4x^2-4={(\sqrt{a}+\frac{1}{\sqrt{a}})}^2-4={(\sqrt{a}-\frac{1}{\sqrt{a}})}^2$

$\Rightarrow 2\sqrt{x^2-1}=\sqrt{a}-\frac{1}{\sqrt{a}}$

Therefore, $\frac{\sqrt{x^2-1}}{x-\sqrt{x^2-1}}=\frac{\sqrt{a}-\frac{1}{\sqrt{a}}}{(\sqrt{a}+\frac{1}{\sqrt{a}})-(\sqrt{a}-\frac{1}{\sqrt{a}})}=\frac{a-1}{2}$