## Precalculus (6th Edition) Blitzer

The simplified form of the expression ${{\left( \frac{a}{c} \right)}^{2}}+{{\left( \frac{b}{c} \right)}^{2}}$ is $1$.
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 $1$.