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. 
 

$$\begin{align}
\sqrt{1-\frac{(m^2+2mn)^2}{(m^2+2mn+2n^2)^2}}&=\sqrt{\frac{(m^2+2mn+2n^2)^2-(m^2+2mn)^2}{(m^2+2mn+2n^2)^2}}\\
&=\sqrt{\frac{(2n(m+n))^2}{(m^2+2mn+2n^2)^2}}\tag{1}\\
&=\frac{2n(m+n)}{m^2+2mn+2n^2}\\
\end{align}$$

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

