Writing 1+1= 2 in a complicated way

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I am learning Unit Circle at the moment and I am using this source as an education tool Trigonometry: Unit Circle (Starts at 20:00).

I understand steps one and two, but when it comes to the third step, where he "cleans up" again, I get lost in thought. I would Google this, but I do not know the technical term, hence the title saying "Writing 1+1= 2 in a complicated way".

Can someone please give me the technical term, for this and simplify the learning curve. Or do I not need this understanding, in order to move on to the next stage where Pythagoras theorem is used to find generated points in a circle and so on.

Given the frames you've posted, it appears the video starts with an equation $$ a + b = c $$ involving positive numbers, and after some algebraic manipulation arrives at a point on the "unit circle": $$ \left(\frac{\sqrt{a}}{\sqrt{c}}\right)^{2} + \left(\frac{\sqrt{b}}{\sqrt{c}}\right)^{2} = 1. $$ I don't know of a specific name for this, but in case it helps, here are the steps: \begin{align*} a + b &= c && \\ (\sqrt{a})^{2} + (\sqrt{b})^{2} &= (\sqrt{c})^{2} && \text{$x = (\sqrt{x})^{2}$ if $x \geq 0$;} \\ \frac{(\sqrt{a})^{2}}{(\sqrt{c})^{2}} + \frac{(\sqrt{a})^{2}}{(\sqrt{c})^{2}} &= 1 && \text{divide through by $(\sqrt{c})^{2}$;} \\ \left(\frac{\sqrt{a}}{\sqrt{c}}\right)^{2} + \left(\frac{\sqrt{b}}{\sqrt{c}}\right)^{2} &= 1 && \frac{(\sqrt{a})^{2}}{(\sqrt{c})^{2}} = \left(\frac{\sqrt{a}}{\sqrt{c}}\right)^{2}. \end{align*} Introducing $$ x = \frac{\sqrt{a}}{\sqrt{c}},\quad y = \frac{\sqrt{b}}{\sqrt{c}}, $$ you have $x^{2} + y^{2} = 1$.

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