If a = $\dfrac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}}$, find the value of :

$\dfrac{a + 2\sqrt{2}}{a - 2\sqrt{2}} + \dfrac{a + 2\sqrt{3}}{a - 2\sqrt{3}}$.

Given,

$\Rightarrow a = \dfrac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}} \\[1em] \Rightarrow \dfrac{a}{2\sqrt{2}} = \dfrac{2\sqrt{3}}{\sqrt{2} + \sqrt{3}}$

Applying componendo and dividendo:

$\Rightarrow \dfrac{a + 2\sqrt{2}}{a - 2\sqrt{2}} = \dfrac{2\sqrt{3} + (\sqrt{2} + \sqrt{3})}{2\sqrt{3} - (\sqrt{2} + \sqrt{3})} \\[1em] \Rightarrow \dfrac{a + 2\sqrt{2}}{a - 2\sqrt{2}} = \dfrac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} …….(i)$

Again,

$\Rightarrow a = \dfrac{4\sqrt{6}}{\sqrt{2} + \sqrt{3}} \\[1em] \Rightarrow \dfrac{a}{2\sqrt{3}} = \dfrac{2\sqrt{2}}{\sqrt{2} + \sqrt{3}}$

Applying componendo and dividendo:

$\Rightarrow \dfrac{a + 2\sqrt{3}}{a - 2\sqrt{3}} = \dfrac{2\sqrt{2} + (\sqrt{2} + \sqrt{3})}{2\sqrt{2} - (\sqrt{2} + \sqrt{3})} \\[1em] \Rightarrow \dfrac{a + 2\sqrt{3}}{a - 2\sqrt{3}} = \dfrac{3\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} …….(ii)$

Adding (i) and (ii) we get,

$\Rightarrow \dfrac{a + 2\sqrt{2}}{a - 2\sqrt{2}} + \dfrac{a + 2\sqrt{3}}{a - 2\sqrt{3}} = \dfrac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} + \dfrac{3\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} \\[1em] = \dfrac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} + \Big(-\dfrac{3\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{2}}\Big) \\[1em] = \dfrac{3\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} - \dfrac{3\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{2}} \\[1em] = \dfrac{3\sqrt{3} + \sqrt{2} - 3\sqrt{2} - \sqrt{3}}{\sqrt{3} - \sqrt{2}} \\[1em] = \dfrac{2\sqrt{3} - 2\sqrt{2}}{\sqrt{3} - \sqrt{2}} \\[1em] = \dfrac{2(\sqrt{3} - \sqrt{2})}{\sqrt{3} - \sqrt{2}} \\[1em] = 2.$

Hence, $\dfrac{a + 2\sqrt{2}}{a - 2\sqrt{2}} + \dfrac{a + 2\sqrt{3}}{a - 2\sqrt{3}}$ = 2.