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

Dec 19, 2017

See below.

#### Explanation:

Using the de Moivre's identity which states

${e}^{i x} = \cos x + i \sin x$ we have

$\frac{1 + {e}^{i x}}{1 + {e}^{- i x}} = {e}^{i x} \frac{1 + {e}^{- i x}}{1 + {e}^{- i x}} = {e}^{i x}$

NOTE

${e}^{i x} \left(1 + {e}^{- i x}\right) = \left(\cos x + i \sin x\right) \left(1 + \cos x - i \sin x\right) = \cos x + {\cos}^{2} x + i \sin x + {\sin}^{2} x = 1 + \cos x + i \sin x$

or

$1 + \cos x + i \sin x = \left(\cos x + i \sin x\right) \left(1 + \cos x - i \sin x\right)$

Dec 19, 2017

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, $z = \frac{1 + \sin x + i \cos x}{1 + \sin x - i \cos x} .$

Multiplying $N r . \mathmr{and} D r .$ by the conjugate of $D r . ,$ we get,

Then, $z = \frac{1 + \sin x + i \cos x}{1 + \sin x - i \cos x} \times \frac{1 + \sin x + i \cos x}{1 + \sin x + i \cos x}$,

$= {\left(1 + \sin x + i \cos x\right)}^{2} / \left\{{\left(1 + \sin x\right)}^{2} - {i}^{2} {\cos}^{2} x\right\}$,

$= {\left(1 + \sin x + i \cos x\right)}^{2} / \left\{{\left(1 + \sin x\right)}^{2} + {\cos}^{2} x\right\}$,

Here, $\text{the Nr.=} {\left(1 + \sin x + i \cos x\right)}^{2} ,$

$= 1 + {\sin}^{2} x - {\cos}^{2} x + 2 \sin x + 2 i \sin x \cos x + 2 i \cos x ,$

$= {\sin}^{2} x + {\sin}^{2} x + 2 \sin x + 2 i \sin x \cos x + 2 i \cos x ,$

$= 2 {\sin}^{2} x + 2 \sin x + 2 i \sin x \cos x + 2 i \cos x ,$

$= 2 \sin x \left(\sin x + 1\right) + 2 i \cos x \left(\sin x + 1\right) ,$

$= 2 \left(\sin x + i \cos x\right) \left(\sin x + 1\right) .$

And, $\text{the Dr.=} {\left(1 + \sin x\right)}^{2} + {\cos}^{2} x$,

$= 1 + 2 \sin x + {\sin}^{2} x + {\cos}^{2} x ,$

$= 1 + 2 \sin x + 1 ,$

$= 2 \sin x + 2 ,$

$= 2 \left(\sin x + 1\right) .$

$\Rightarrow z = \frac{2 \left(\sin x + i \cos x\right) \left(\sin x + 1\right)}{2 \left(\sin x + 1\right)}$,

$= \sin x + i \cos x .$

Q.E.D.

Enjoy Maths.!