If $\frac{\sqrt{5}-1}{\sqrt{5}+1}=a+b\sqrt{5}$, then find the value of $a$ and $b$.

Let's rationalize the denominator on the left-hand side:
$$\frac{\sqrt{5}-1}{\sqrt{5}+1}\times \frac{\sqrt{5}-1}{\sqrt{5}-1}=\frac{(\sqrt{5}-1{)}^2}{(\sqrt{5}{)}^2-(1{)}^2}$$
Expanding the numerator using $(x-y{)}^2=x^2-2xy+y^2$:
$$(\sqrt{5}-1{)}^2=5-2\sqrt{5}+1=6-2\sqrt{5}$$
Denominator:
$$5-1=4$$
So, the expression becomes:
$$\frac{6-2\sqrt{5}}{4}=\frac{6}{4}-\frac{2\sqrt{5}}{4}=\frac{3}{2}-\frac{1}{2}\sqrt{5}$$
Comparing this with $a+b\sqrt{5}$:
$$a=\frac{3}{2}$$
$$b=-\frac{1}{2}$$