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i have a problem in dis sum from complex numbers

prove that
(1+sinA+icosA)/(1-sinA-icosA)=i(secA+tanA)

1 answer

multiply the left side by (1-sinA+icosA)/(1-sinA+icosA)
expand and simply it down to

icosA/(1+sinA)

multiply that by (1-sinA)/(1-sinA)
expand again and simplify down to

i(1-sinA)/cosA

= i(1/cosA - sinA/cosA)
= i(secA - tanA)
= Right Side