In $\Delta ABC,(a-b^2){\cos }^2\frac{C}{2}+{(a+b)}^2{\sin }^2\frac{C}{2}=$

${(a-b)}^2{\cos }^2\frac{C}{2}+{(a+b)}^2{\sin }^2\frac{C}{2}$

${(a-b)}^2{\cos }^2\frac{C}{2}+{(a+b)}^2\left(1-{\cos }^2\frac{{\displaystyle C}}{{\displaystyle 2}}\right)$

$={(a-b)}^2{\cos }^2\frac{C}{2}+{(a+b)}^2-{(a+b)}^2{\cos }^2\frac{C}{2}$

$={\cos }^2\frac{C}{2}{\left[{(a-b)}^2-(a+b)\right]}^2+{(a+b)}^2$

$={\cos }^2\frac{C}{2}[-4ab]+{(a+b)}^2$

$=\frac{s(s-c)}{ab}(-4ab)+{(a+b)}^2$

$\begin{array}{l}={(a+b)}^2-4s(s-c)\\ =a^2+b^2+2ab-4(s^2-sc)\end{array}$

$=a^2+b^2+2ab-4\left[{\left(\frac{{\displaystyle a+b+c}}{{\displaystyle 2}}\right)}^2-\left(\frac{{\displaystyle a+b+c}}{{\displaystyle 2}}\right)c\right]$

$=a^2+b^2+2ab-4\left[\frac{{(a+b+c)}^2-2(a+b+c)c}{4}\right]$

$\begin{array}{l}=a^2+b^2+2ab-a^2-b^2-c^2-2ab-2bc-2ca+2ac+2bc+2c^2\\ =c^2\end{array}$