We have

$sec(\alpha + i\beta) = \dfrac{1}{cos(\alpha + i\beta)}$

                 =$ \dfrac{1}{cos\alpha cosi\beta + sin\alpha sini\beta}$

                =$ \dfrac{1}{cos\alpha h\beta + isin\alpha sinh\beta}$

                = $\dfrac{1}{(\dfrac{x}{2}) + i(\dfrac{y}{2})}$………………{by given data}

                = $\dfrac{2}{x+iy}$..................................(1)

Similarly

$sec(\alpha – i\beta) =\dfrac{1}{cos(\alpha – i\beta)}$

               =$ \dfrac{1}{cos\alpha cosi\beta – sin\alpha sini\beta}$

              =$ \dfrac{1}{ cos\alpha cosh\beta – sin\alpha sinh\beta }$

              =$ \dfrac{1}{(\dfrac{x}{2}) - i(\dfrac{y}{2})}$………………{by given data}

              = $\dfrac{2}{x-iy}$.........................(2)

$\therefore sec(\alpha – i\beta) + sec(\alpha + i\beta) = \dfrac{2}{x-iy} + \dfrac{2}{x+iy} $......................(from equation 1 and 2)

                                       =$ \dfrac{2(x+iy) +2(x-iy)}{x^{2}-y^{2}}$

                                       =$\dfrac{2x+2iy+2x-2iy}{x^{2}-y^{2}}$…………………$(\because i =(-1) ) $

                                       =$\dfrac{4x}{x^{2}+y{2}} $     

$\therefore sec(\alpha – i\beta) + sec(\alpha + i\beta) =\dfrac{4x}{x^{2}+y{2}}$