Question

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

Answer 1

See below.

Explanation:

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

Kindly refer to a Proof in The Explanation.

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.!