Note: Not really the recommended justification here. Firstly (and somewhat less importantly) note that saying \( \sqrt{a^2} = \pm a\) is technically incorrect, because that means you're claiming that \( \sqrt{a^2} \) could potentially equal two different values. This would be problematic when saying something like \( \sqrt{5^2} = \pm 5 \). In general, the safest bet is to always say that \( \sqrt{a^2} = |a| \) when doing these problems correctly.

But it's less of an issue here because you attempted to justify your way around it.

The justification behind the "\(\pm\)'s cancelling out into a negatives" isn't sufficient. The subtlety is that you haven't convinced me between which "\( \sqrt{\frac{a^2}{b^2}}=\frac{\sqrt{a^2}}{\sqrt{b^2}}=\frac{\pm a}{\pm b} \)" or "\( \sqrt{\frac{a^2}{b^2}}=\frac{\sqrt{a^2}}{\sqrt{b^2}}=\frac{\pm a}{\mp b} \)" is the favourable one. Saying that "observing that taking the negative one for the top one will give the desired answer" is not convincing because it looks like you're assuming what you're trying to prove,

A better justification here would be that \( \sqrt{\frac{(\sin\theta-\cos\theta)^2}{(\sin\theta+\cos\theta)^2}} = \left| \frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta} \right| = \left| \frac{\tan\theta-1}{\tan\theta+1} \right| = -\frac{\tan\theta-1}{\tan\theta+1} \), justifying the last equality here by the fact that \( \boxed{0\leq\theta\leq\frac\pi4} \).

We know that over this domain, \(0 \leq \tan\theta \leq 1\). Hence on one hand, it is clear that \(\tan\theta+1 > 0\). On the other hand, rearranging \(\tan\theta \leq 1\) gives \(\tan\theta - 1\leq 0\). Therefore \(\frac{\tan\theta-1}{\tan\theta+1} \leq0\) (non-positive number divided by positive number), so \(\left|\frac{\tan\theta-1}{\tan\theta+1}\right| = -\frac{\tan\theta-1}{\tan\theta+1}\). This now relies on the actual definition of the absolute value

\[ |x| = \begin{cases}x&x\geq 0\\ -x&x<0\end{cases} \]

instead of intuition that cannot be justified.