Correct Answer: $\frac{1}{2}(a–1)$

Solution : Given: $2x=\sqrt{a}+\frac{1}{\sqrt{a}}, a>0$
$⇒2x=\sqrt{a}+\frac{1}{\sqrt{a}}$
$⇒x=\frac{1}{2}(\sqrt{a}+\frac{1}{\sqrt{a}})$
Calculating the value of ${\sqrt{x^{2}-1}}$
$⇒{\sqrt{x^{2}-1}}={\sqrt{(\frac{1}{2}(\sqrt{a}+\frac{1}{\sqrt{a}}))^{2}-1}}$
$⇒{\sqrt{(\frac{a^2-2a+1}{4a})}}=\frac{a-1}{2\sqrt{a}}$
So that $\frac{\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}$ becomes
$⇒\frac{\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}=\frac{\frac{a-1}{2\sqrt{a}}}{\frac{1}{2}(\sqrt{a}+\frac{1}{\sqrt{a}})-(\frac{a-1}{2\sqrt{a}})}$
$⇒\frac{\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}=\frac{a-1}{2\sqrt{a}}×\frac{2\sqrt{a}}{2}=\frac{1}{2}(a–1)$
Hence, the correct answer is $\frac{1}{2}(a–1)$.