Rationalise the denominator of $ \frac{\sqrt{a}+1}{\sqrt{a}-1} $.

Given:

\( \frac{\sqrt{a}+1}{\sqrt{a}-1} \).

To do: 

We have to rationalise the denominator of the given expression.

Solution:

We know that,

Rationalising factor of a fraction with denominator ${\sqrt{a}-\sqrt{b}}$ is ${\sqrt{a}+\sqrt{b}}$.

Therefore,

Rationalising factor of a fraction with denominator ${\sqrt{a}-1}$ is ${\sqrt{a}+1}$.

This implies,

$\frac{\sqrt{a}+1}{\sqrt{a}-1}=\frac{\sqrt{a}+1}{\sqrt{a}-1}\times\frac{\sqrt{a}+1}{\sqrt{a}+1}$

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

$=\frac{(\sqrt{a})^2+2\times \sqrt{a}\times1+(1)^2}{a-1}$

$=\frac{a+2\sqrt{a}+1}{a-1}$.