Question

Prove that :$\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$

Answer 1

$\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{(\sqrt{3}-1{)}^2}{(\sqrt{3}{)}^2-1^2}=\frac{(\sqrt{3}{)}^2+(1{)}^2-2\sqrt{3}}{3-1}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$