\(x = \cfrac{1}{2 + \sqrt{3}}\) \(= \cfrac{(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})}\) \(= \cfrac{(2 - \sqrt{3})}{4 - 3}\) \(= 2 - \sqrt{3}\) \(\therefore 1 + x = 1 + 2 - \sqrt{3} = 3 - \sqrt{3}\) Again, \(y = \cfrac{1}{2 - \sqrt{3}}\) \(= \cfrac{(2 + \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})}\) \(= \cfrac{(2 + \sqrt{3})}{4 - 3}\) \(= 2 + \sqrt{3}\) \(\therefore 1 + y = 1 + 2 + \sqrt{3} = 3 + \sqrt{3}\) \(\cfrac{1}{1 + x} + \cfrac{1}{1 + y}\) \(= \cfrac{1}{3 - \sqrt{3}} + \cfrac{1}{3 + \sqrt{3}}\) \(= \cfrac{(3 + \sqrt{3}) + (3 - \sqrt{3})}{(3 - \sqrt{3})(3 + \sqrt{3})}\) \(= \cfrac{6}{9 - 3}\) \(= \cfrac{6}{6} = 1\) (Proved)