written 3.9 years ago by |
We have
sec(α+iβ)=1cos(α+iβ)
=1cosαcosiβ+sinαsiniβ
=1cosαhβ+isinαsinhβ
= 1(x2)+i(y2)………………{by given data}
= 2x+iy..................................(1)
Similarly
sec(α–iβ)=1cos(α–iβ)
=1cosαcosiβ–sinαsiniβ
=1cosαcoshβ–sinαsinhβ
=1(x2)−i(y2)………………{by given data}
= 2x−iy.........................(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}}