Answer
True for $a\ne1$.
Work Step by Step
This is mostly true. We can show that the left equation turns into the right equation by multiplying the numerator and denominator by $1-\sqrt{a}$ and using the fact that $(a-b)(a+b)=a^2-b^2$:
$\displaystyle\frac{1+\sqrt{a}}{1-a}=\frac{1+\sqrt{a}}{1-a}\cdot\frac{1-\sqrt{a}}{1-\sqrt{a}}=\frac{(1-a)}{(1-a)(1-\sqrt{a})}=\frac{1}{1-\sqrt{a}}$
However, $a$ can not be $1$ because then the denominator would be $0$ and the equation would be undefined. Thus the equation is true for $a\ne1$.
