# Complex numbers question. Help please?

## Prove that ${\left(1 - \cos x + 2 i \cos x\right)}^{- 1} = \frac{1 - 2 i \cot \left(\frac{x}{2}\right)}{5 + 3 \cos x}$? So you don't have to retype the equation, without the hashtags it looks like this: (1-cosx+2icosx)^(-1)=(1-2icot(x/2))/(5+3cosx)

Oct 15, 2017

Please refer to the Explanation. Kindly, check the Problem.

#### Explanation:

If we substitute $x = \frac{\pi}{2} ,$ then,

$\text{The L.H.S.=} {\left(1 - \cos x + 2 i \cos x\right)}^{-} 1 ,$

$= {\left\{1 - \cos \left(\frac{\pi}{2}\right) + 2 i \cos \left(\frac{\pi}{2}\right)\right\}}^{-} 1 ,$

$= {\left\{1 - 0 + 2 i \left(0\right)\right\}}^{-} 1 ,$

$= {1}^{-} 1 ,$

$= 1 ,$ while, for, $x = \frac{\pi}{2} ,$

$\text{The R.H.S.=} \frac{1 - 2 i \cot \left(\frac{x}{2}\right)}{5 + 3 \cos x} ,$

$= \frac{1 - 2 i \cot \left(\frac{\frac{\pi}{2}}{2}\right)}{5 + 3 \cos \left(\frac{\pi}{2}\right)} ,$

$= \frac{1 - 2 i \cot \left(\frac{\pi}{4}\right)}{5 + 3 \left(0\right)} ,$

$= \frac{1 - 2 i \cdot 1}{5} ,$

$= \frac{1 - 2 i}{5.}$

$\text{Therefore, The L.H.S."!="The R.H.S.}$

Kindly, check the Problem.