Question

If $\frac{\sqrt{3}-1}{\sqrt{3}+1}=a-b\sqrt{3}$, then

• A. $a=2,b=1$
• B. $a=2,b=-1$
• C. $a=-2,b=1$
• D. $a=b=1$

Answer 1

The correct option is A

$a=2,b=1$

$\frac{\sqrt{3}-1}{\sqrt{3}+1}=a-b\sqrt{3}$

$\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}$

(Rationalising the denominator)

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

$=2-\sqrt{3}$

Now, $2-\sqrt{3}=a-b\sqrt{3}$

Comparing, we get

a = 2, b= 1