Question

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Answer 1

QUESTION:-

$\bf\;\frac{(\sqrt{3}-1)}{(\sqrt{3}+1)}=a-b\sqrt{3}$

EXPLANATION:-

How to solve ??

First we will simplify RHS by rationalizing denominator

So first let's rationalize RHS

So to rationlaize the R.H.S  we will multiply it by the conjugate of it's denominator i.e. √3-1

$\bf\;\frac{(\sqrt{3}-1)}{(\sqrt{3}+1)}\times\frac{\sqrt{3}-1}{\sqrt{3}-1}\\\\\frac{(\sqrt{3}-1)^2}{((\sqrt{3})^2-(1)^2} \;\;\;\sf\;using\;(a+b)(a-b)=a^2-b^2\\\\\frac{3+1+2\sqrt{3}}{2} \;\;\;using\;(a-b)^2=a^2+b^2+2ab\\\\\frac{4+2\sqrt{3}}{2} \\\\\frac{2(2+\sqrt{3})}{2} \;\;\;(taking\;2\;common\;in\;numerator)\\\\2+\sqrt{3}\;\;\;\;(cancel\;2\;in\;numerator\;and\;denominator)$

Henceforth the denominator is rationalized

Now let's compare it

$\bf\;2+\sqrt{3}=a-b\sqrt{3}\\\\\\We\;can\;alse;write\;2+\sqrt{3}\;as\\\\\\2-(-1\sqrt{3})\\\\\\\Now\;let's\;compare\;\\\\\\2-(-1\sqrt{3})=a-b\sqrt{3}\\\\\\On\;comparing\;we\;get:-\\\\\boxed{\bf\;a=2\;and\;b=1}$