If $\dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} = a + b\sqrt{3}$ ,find the values of 'a' and 'b'.

Given,

Equation : $\dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} = a + b\sqrt{3}$

Rationalizing the denominator of L.H.S. of the above equation :

$\Rightarrow \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} \\[1em] \Rightarrow \dfrac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} \\[1em] \Rightarrow \dfrac{(\sqrt{3})^2 + 1^2 + 2 \times \sqrt{3} \times 1}{3 - 1} \\[1em] \Rightarrow \dfrac{3 + 1 + 2\sqrt{3}}{2} \\[1em] \Rightarrow \dfrac{4 + 2\sqrt{3}}{2} \\[1em] \Rightarrow \dfrac{2(2 + \sqrt{3})}{2} \\[1em] \Rightarrow 2 + \sqrt{3}.$

Comparing $2 + \sqrt{3} \text{ with } a + b\sqrt{3}$, we get :

a = 2 and b = 1.

Hence, a = 2 and b = 1.