Question

In $\Delta ABC{\left(a-b\right)}^{2}co{s}^2\frac{C}{2}+{\left(a+b\right)}^{2}si{n}^2\frac{C}{2}=$

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

Answer 1

Answer: (B)

$(a-b{)}^{2}co{s}^2\frac{C}{2}+(a+b{)}^{2}si{n}^2\frac{C}{2}$
$=\left(a^2+b^2-2ab\right)co{s}^2\frac{C}{2}+a^2+b^2+2absi{n}^2\frac{C}{2}$
$=a^2\left(co{s}^2\frac{C}{2}+si{n}^2\frac{C}{2}\right)+b^2\left(co{s}^2\frac{C}{2}+{\sin }^2\frac{C}{2}\right)$
$-2ab\left(co{s}^2\frac{C}{2}-si{n}^2\frac{C}{2}\right)$
$=a^2+b^2-2abcosC$
$\left[\because co{s}^2\frac{C}{2}-{\sin }^2\frac{C}{2}=\cos C\right]$
$=a^2+b^2-2ab\left(\frac{a^2+b^2-c^2}{2ab}\right)=c^2$
$\left[\because \cos C=\frac{a^2+b^2}{2ab}\right]$