Proving complex phasor expression equals (1 - cos θ) / sin θ

[Loudhvx]

Please forgive me as I may have to edit this post to get the equations to show properly.

I am doing some work with AC circuits and part of one of my phasor equations has this in it:
$\frac {2i} {1+cos(θ) + i sin(θ)} - i$,
where i is the imaginary number $\sqrt{-1}$.

However, knowing the physics of the circuit tells me the end result should have no complex numbers.
This means "i" should drop out of this part of the equation.

When plugging in values for $\theta$, the i terms do in fact drop out.

By plugging in numbers, I found the result is always of the form:
$\frac{a}{b} ( \frac{ai + b}{b + ai})$,
where $a = 1 - cos (\theta)$,
and $b = sin (\theta)$.
The portion in the parenthesis, obviously, always equals 1.

This gives me the result:
$\frac {1 - cos (\theta)} {sin (\theta)}$,
which satisfies the requirement of having no imaginary numbers.

Using this in my general equation seems to work as I confirmed with certain specific cases for which the solution was verifiable by other means.

My question, or my request for assistance, is in getting a formal proof that for all angles:
$\frac {2i} {1+cos(θ) + i sin(θ)} - i = \frac {1 - cos (\theta)} {sin (\theta)}$.

I tried using some trig identities and even Euler's identity, but, unfortunately, my trig is a little rusty.

Thank you for your time in advance.

 

[SteamKing]

The math typing looks OK; what is you question?

 

[Loudhvx]

I finished editing the post above, with my question.

 

[Saitama]

Multiply and divide by $1+\cos\theta-i\sin\theta$.

$$\frac{2i}{1+\cos\theta+i\sin\theta}-i=\frac{2i(1+\cos\theta-i\sin\theta)}{(1+\cos\theta)^2+\sin^2\theta}-i=\frac{i(1+\cos\theta-i\sin\theta)}{1+\cos\theta}-i$$
$$=i+\frac{\sin\theta}{1+\cos\theta}-i=\frac{\sin\theta}{1+\cos\theta}\times \frac{1-\cos\theta}{1-\cos\theta}=\frac{1-\cos\theta}{\sin\theta}$$

 

[Loudhvx]

Thank you very very much! :)

and I guess I put this in the wrong forum. Sorry about that.