Simple Trigonometric Problem Here is a Trig problem, and I am missing some understanding of some basic algebraic rule:
$if \sin\theta= \frac{m^2 +2mn}{m^2+2mn+2n^2}$
then prove that $\tan\theta = \frac {m^2 +2mn}{2mn+2n^2},$ In the picture below can anyone explain me how did they achieve the step from 1 to 2 highlighted in red. 
 
 A: \begin{align*}
\sqrt{1-\left(\frac{m^2+2mn}{m^2+2mn+2n^2}\right)^2} 
&= \sqrt{\frac{(m^2+2mn+2n^2)^2 - (m^2+2mn)^2}{(m^2+2mn+2n^2)^2}}\\
&= \sqrt{\frac{4(m^2+2mn)2n^2 + 4n^4}{(m^2+2mn+2n^2)^2}}\\
&= \sqrt{\frac{4n^2(m^2+2mn+n^2)}{(m^2+2mn+2n^2)^2}}\\
&= \sqrt{\frac{4n^2(m+n)^2}{(m^2+2mn+2n^2)^2}}\\
&= \sqrt{\left(\frac{2n(m+n)}{m^2+2mn+2n^2}\right)^2}\\
&= \frac{2n(m+n)}{m^2+2mn+2n^2}
\end{align*}
A: Hint: $$\begin{align}
1-\left(\frac{m^2+2mn}{m^2+2mn+2n^2}\right)^2
&=\left[1\color{red}{+}\frac{m^2+2mn}{m^2+2mn+2n^2}\right]\cdot\left[1\color{red}{-}\frac{m^2+2mn}{m^2+2mn+2n^2}\right]\\
&=\left[\frac{2m^2+4mn+2n^2}{m^2+2mn+2n^2}\right]\cdot\left[\frac{2n^2}{m^2+2mn+2n^2}\right]\\
\end{align}$$
A: $$\begin{align}
\sqrt{1-\frac{(m^2+2mn)^2}{(m^2+2mn+2n^2)^2}}&=\sqrt{\frac{(m^2+2mn+2n^2)^2-(m^2+2mn)^2}{(m^2+2mn+2n^2)^2}}\\
&=\sqrt{\frac{(2n(m+n))^2}{(m^2+2mn+2n^2)^2}}\tag{1}\\
&=\frac{2n(m+n)}{m^2+2mn+2n^2}\\
\end{align}$$

$\text{Explanation } 1$ $$(a+b)^2-a^2 = 2ab+b^2=b(2a+b)$$
  Where $a=m^2+2mn$ , $b=2n^2$

A: we have $(m^2+2mn+2n^2)^2-(m^2+2mn)^2=(m^2+2mn+2n^2-m^2-2mn)(m^2+2mn+2n^2+m^2+2mn)=4m^2(m+n)^2$
