Answer
The simplified form of the expression ${{\left( \frac{a}{c} \right)}^{2}}+{{\left( \frac{b}{c} \right)}^{2}}$ is $1$.
Work Step by Step
Consider the expression, ${{\left( \frac{a}{c} \right)}^{2}}+{{\left( \frac{b}{c} \right)}^{2}}$
$\begin{align}
& {{\left( \frac{a}{c} \right)}^{2}}+{{\left( \frac{b}{c} \right)}^{2}}=\frac{{{a}^{2}}}{{{c}^{2}}}+\frac{{{b}^{2}}}{{{c}^{2}}} \\
& =\frac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}}
\end{align}$
The Pythagorean identity is ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$.
Substitute ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$ in the expression $\frac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}}$.
$\begin{align}
& {{\left( \frac{a}{c} \right)}^{2}}+{{\left( \frac{b}{c} \right)}^{2}}=\frac{{{c}^{2}}}{{{c}^{2}}} \\
& =1
\end{align}$
Therefore, the simplified form of the expression ${{\left( \frac{a}{c} \right)}^{2}}+{{\left( \frac{b}{c} \right)}^{2}}$ is $1$.
