<s>[INST] 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. 
 
 [/INST] \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*}
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