hello

here's my function :

let $\displaystyle \mathbb{E} = [-\pi ,+\pi]$

$\displaystyle x \in \mathbb{E}$ $\displaystyle ; $$\displaystyle f(x) = \sqrt{\frac{1+cos(x)}{1-cos(x)}}$

and here's my questions :

Prove that $\displaystyle (\forall x \in \mathbb{D})$ $\displaystyle f(x) = \left |\tan (\frac{\pi}{2}-\frac{x}{2}) \right |$

Study the sign of $\displaystyle \tan (\frac{\pi}{2}-\frac{x}{2})$ in $\displaystyle ]0,+\pi]$

i proceeded like this :

for the first one,we have $\displaystyle \forall x\in \mathbb{R},1+cos(x) = 2cos^{2}(\frac{x}{2})$ and $\displaystyle 1-cos(x) = 2sin^{2}(\frac{x}{2})$

$\displaystyle \Rightarrow \sqrt{\frac{1+cos(x)}{1-cos(x)}} =$ $\displaystyle \sqrt{\frac{2cos^{2} (\frac{x}{2})}{2sin^{2}(\frac{x}{2})}}=$

$\displaystyle | \frac{cos(\frac{x}{2})}{sin(\frac{x}{2})}| $ $\displaystyle = \left |\frac{1}{\tan (\frac{x}{2})} \right |$ $\displaystyle = \left |\tan (\frac{\pi}{2}-\frac{x}{2}) \right |$

for the second one,am i correct writing ; $\displaystyle \forall x \in \mathbb{D},f(x) = tan(\frac{\pi}{2}-\frac{x}{2})$ ( i already studied $\displaystyle f$ )

thanks for everyone

by the way i'm trying to improve myself for writing proofs so if there is any small mistake please tell me