Prove that :

$\Big(\dfrac{\text{tan 60° + 1}}{\text{tan 60° - 1}}\Big)^2 = \dfrac{\text{1 + cos 30°}}{\text{1 - cos 30°}}$

$\Big(\dfrac{\text{tan 60° + 1}}{\text{tan 60° - 1}}\Big)^2 = \dfrac{\text{1 + cos 30°}}{\text{1 - cos 30°}}$

$\text{L.H.S.} = \Big(\dfrac{\text{tan 60° + 1}}{\text{tan 60° - 1}}\Big)^2\\[1em] = \Big(\dfrac{\sqrt3 + 1}{\sqrt3 - 1}\Big)^2\\[1em] = \Big(\dfrac{(\sqrt3 + 1) \times (\sqrt3 + 1)}{(\sqrt3 - 1) \times (\sqrt3 + 1)}\Big)^2\\[1em] = \Big(\dfrac{(\sqrt3 + 1)^2}{(\sqrt3)^2 - (1)^2}\Big)^2\\[1em] = \Big(\dfrac{3 + 1 + 2 \times 1 \times \sqrt3}{3 - 1}\Big)^2\\[1em] = \Big(\dfrac{4 + 2\sqrt3}{2}\Big)^2\\[1em] = (2 + \sqrt3)^2\\[1em] = 4 + 3 + 2 \times 2 \times \sqrt3\\[1em] = 7 + 4\sqrt3$

$\text{R.H.S.} = \dfrac{\text{1 + cos 30°}}{\text{1 - cos 30°}}\\[1em] = \dfrac{1 + \dfrac{\sqrt3}{2}}{1 - \dfrac{\sqrt3}{2}}\\[1em] = \dfrac{\dfrac{2 + \sqrt3}{2}}{\dfrac{2 - \sqrt3}{2}}\\[1em] = \dfrac{\dfrac{2 + \sqrt3}{\cancel{2}}}{\dfrac{2 - \sqrt3}{\cancel{2}}}\\[1em] = \dfrac{2 + \sqrt3}{2 - \sqrt3}\\[1em] = \dfrac{(2 + \sqrt3) \times (2 + \sqrt3)}{(2 - \sqrt3) \times (2 + \sqrt3)}\\[1em] = \dfrac{(2 + \sqrt3)^2}{(2)^2 - (\sqrt3)^2}\\[1em] = \dfrac{4 + 3 + 2 \times 2 \times \sqrt3}{4 - 3}\\[1em] = 7 + 4\sqrt3\\[1em]$

∴ L.H.S. = R.H.S.

Hence, $\Big(\dfrac{\text{tan 60° + 1}}{\text{tan 60° - 1}}\Big)^2 = \dfrac{\text{1 + cos 30°}}{\text{1 - cos 30°}}$