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If cos α cosh \( \beta =\frac{x}{2},\sin \alpha sinh \beta =\frac{y}{2}, \)/ Prove that \( sec\left ( \alpha -i\beta \right )+sec\left ( \alpha +i\beta \right )= \frac{4x}{x^2+y^2}\)/
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We have

sec(α+iβ)=1cos(α+iβ)

                 =1cosαcosiβ+sinαsiniβ

                =1cosαhβ+isinαsinhβ

                = 1(x2)+i(y2)………………{by given data}

                = 2x+iy..................................(1)

Similarly

sec(αiβ)=1cos(αiβ)

               =1cosαcosiβsinαsiniβ

              =1cosαcoshβsinαsinhβ

              =1(x2)i(y2)………………{by given data}

              = 2xiy.........................(2)

......................(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}}

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