Question

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

• A. $b^2$
• B. $c^2$
• C. $a^2$
• D. $a^2+b^2+c^2$

Answer 1

The correct option is B $c^2$

${(a-b)}^2{cos}^2\frac{C}{2}+{(a+b)}^2{sin}^2\frac{C}{2}=a^2+b^2+2ab({sin}^2\frac{C}{2}-{cos}^2\frac{C}{2})=a^2+b^2+2ab(2{sin}^2\frac{C}{2}-1)$

But $sin\frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{ab}}$

where $s=\frac{a+b+c}{2}$

$\therefore$ $a^2+b^2+2ab(2{sin}^2\frac{C}{2}-1)=a^2+b^2+2ab(\frac{(c+b-a)(c+a-b)}{2ab}-1)=a^2+b^2+c^2-{(b-a)}^2-2ab=c^2$