What is the value of ( a - b ) 2 cos 2 c 2 + ( a + b ) 2 sin 2 c 2 =

Given that: $(a-b{)}^2{\cos }^2\frac{c}{2}+(a+b{)}^2{\sin }^2\frac{c}{2}$

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

$=(a^2+b^2\left)\right({\cos }^2\frac{c}{2}+{\sin }^2\frac{c}{2})+2ab({\sin }^2\frac{c}{2}-{\cos }^2\frac{c}{2})$

As we know that $\cos 2A={\cos }^{2}A-{\sin }^{2}A$

$=(a^2+b^2)-2ab\cos C$

As we know that $\cos C=\frac{a^2+b^2-c^2}{2ab}$

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

$=c^2$