Prove #(1+sinx+icosx)/(1+sinx-icosx)=sinx+icosx#?

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Answer 1 · 2017-12-19T01:31:18

See below.

Using the de Moivre's identity which states

#e^(ix) = cos x+i sin x# we have

#(1+e^(ix))/(1+e^(-ix)) = e^(ix)(1+e^(-ix))/(1+e^(-ix)) = e^(i x)#

NOTE

#e^(ix)(1+e^(-ix)) = (cos x+isinx)(1+cosx-i sinx) = cosx+cos^2x+isinx+sin^2x = 1+cosx +isinx#

or

#1+cosx +isinx = (cos x+isinx)(1+cosx-i sinx) #

Answer 2 · 2017-12-19T12:16:24

Kindly refer to a Proof in The Explanation.

No doubt that Respected Cesareo R. Sir's Answer is the

easiest & shortest one, but, here is another way to solve it :

Let, #z=(1+sinx+icosx)/(1+sinx-icosx).#

Multiplying #Nr. and Dr.# by the conjugate of #Dr.,# we get,

Then, #z=(1+sinx+icosx)/(1+sinx-icosx)xx(1+sinx+icosx)/(1+sinx+icosx)#,

#=(1+sinx+icosx)^2/{(1+sinx)^2-i^2cos^2x}#,

#=(1+sinx+icosx)^2/{(1+sinx)^2+cos^2x}#,

Here, #"the Nr.="(1+sinx+icosx)^2,#

#=1+sin^2x-cos^2x+2sinx+2isinxcosx+2icosx,#

#=sin^2x+sin^2x+2sinx+2isinxcosx+2icosx,#

#=2sin^2x+2sinx+2isinxcosx+2icosx,#

#=2sinx(sinx+1)+2icosx(sinx+1),#

#=2(sinx+icosx)(sinx+1).#

And, #"the Dr.="(1+sinx)^2+cos^2x#,

#=1+2sinx+sin^2x+cos^2x,#

#=1+2sinx+1,#

#=2sinx+2,#

#=2(sinx+1).#

#rArr z={2(sinx+icosx)(sinx+1)}/{2(sinx+1)}#,

#=sinx+icosx.#

Q.E.D.

Enjoy Maths.!