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It seems really strange that $\pm$ can be eliminated, as an algebraic manipulation... it doesn't seem like an algebraic rule.

I think it's as simple as, if $x\ge0$, then $x = \sqrt{x^2}$, and if $x\le0$, then $x = -\sqrt{x^2}$, so we don't need the gosh-I-don't-know $\pm$, but this doesn't really make sense to me.

Just for context, this came up in trig. I worked out the derivation, but am puzzled at the step in the title.

\begin{align} \tan \frac{\theta}{2} &= \pm \sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\\ &= \frac{1-\cos\theta}{\sin\theta} \end{align}

I'm missing something more basic than trig, but I don't recall this ever coming up before. Thanks for any clarity!

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  • $\begingroup$ √x >0 always. The plus/minus sign arises here as tan(x/2) can be positive/negative according to which quadrant it lies. $\endgroup$ – user600016 May 16 at 8:00
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In general the $\pm$ can't be eliminated, as you've already noted. What the derivation should have said is $$\frac{1-\cos\theta}{\sin\theta}=\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}=\tan\frac{\theta}{2}.$$

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