Posted by sonia on .
i have a problem in dis sum from complex numbers
prove that
(1+sinA+icosA)/(1sinAicosA)=i(secA+tanA)

algebra 
Reiny,
multiply the left side by (1sinA+icosA)/(1sinA+icosA)
expand and simply it down to
icosA/(1+sinA)
multiply that by (1sinA)/(1sinA)
expand again and simplify down to
i(1sinA)/cosA
= i(1/cosA  sinA/cosA)
= i(secA  tanA)
= Right Side