How was the plus-minus sign removed in these steps for proving a half-angle identity for tangent?

Question

How was the plus-minus sign removed in the following solution? (see red rectangle)

Answer 1

The derivation is indeed faulty. You can make it rigorous by avoiding the square roots: \begin{align} \tan^2\frac{\theta}{2} &=\frac{1-\cos\theta}{1+\cos\theta} \\[6px] &=\frac{1-\cos^2\theta}{(1+\cos\theta)^2} \\[6px] &=\frac{\sin^2\theta}{(1+\cos\theta)^2} \end{align} Therefore we have $$ \Bigl\lvert\tan\frac{\theta}{2}\Bigr\rvert=\frac{\lvert\sin\theta\rvert}{1+\cos\theta} $$ (note that $1+\cos\theta>0$, so no absolute value is needed in the denominator) and now we can observe that $\tan(\theta/2)$ and $\sin\theta$ have the same sign for every $\theta$ and so the absolute values can be removed.

In a different way: $$ \frac{\sin2\alpha}{1+\cos2\alpha}=\frac{2\sin\alpha\cos\alpha}{1+2\cos^2\alpha-1}=\frac{\sin\alpha}{\cos\alpha}=\tan\alpha $$ Now set $\alpha=\theta/2$.