How do I derive the formula for the reciprocal of a hypotenuse?

Question

Given $a^2 + b^2 = c^2$,

Why is it that the equation below yields the reciprocal of the hypotenuse c, ($\frac{1}c$)?

$\sqrt{(\frac{a}{a^2+b^2})^2 + (\frac{b}{a^2+b^2})^2}$

Worked example:

$3^2 + 4^2 = c^2$
$c = 5$

$\sqrt{(\frac{3}{3^2+4^2})^2 + (\frac{4}{3^2+4^2})^2}$
= $\sqrt{(\frac{3}{25})^2 + (\frac{4}{25})^2}$
= $\sqrt{(.0144) + (.0256)}$
= $\sqrt{.04}$
= .2 or $1/5$

Answer 1

Observe that, for $ab\ne0$, $$ \left(\frac{a}{a^2+b^2}\right)^2 + \left(\frac{b}{a^2+b^2}\right)^2=\frac{a^2}{(a^2+b^2)^2}+\frac{b^2}{(a^2+b^2)^2}=\frac{a^2+b^2}{(a^2+b^2)^2}=\frac{1}{a^2+b^2}. $$ Hope you can take it from here.

Answer 2

Expand the equation: $$ \begin{align} \sqrt{\left(\frac{a}{a^2+b^2}\right)^2+\left(\frac{b}{a^2+b^2}\right)^2}&=\sqrt{\left(\frac{a}{c^2}\right)^2+\left(\frac{b}{c^2}\right)^2}\\ &=\sqrt{\frac{a^2}{c^4}+\frac{b^2}{c^4}}\\ &=\sqrt{\frac{c^2}{c^4}}\\ &=\sqrt{\frac{1}{c^2}}\\ &=\frac{1}{c} \end{align} $$