I’ve got to proof that:
$\tan\left(\frac{A}{2}\right) = \sqrt{\frac{1 -\cos(A)}{1 + \cos(A)}}$
My attempt was (tried with right side):
$= \pm \sqrt{\frac{1 -\cos(A)}{1 + \cos(A)} \cdot \frac{1 -\cos(A)}{1 - \cos(A)}}$ $= \pm \sqrt{\frac{\left(1 -\cos(A)\right)^2}{1 - \cos^2(A)}}$ $= \pm \frac{1 -\cos(A)}{\sin(A)}$
They are not even similar.
I tried Wolfram Alpha online calculator and it showed one of the alternate answers as:
$\sqrt{\tan^2\left(\frac{A}{2}\right)}$
That’s the answer I suppose I should get to, but I’ve tried it many times and I can’t find (imagine) a possible route.
Please, if you could point me in the right direction, I’d greatly appreciate it.
Thank you very much in advance.