Question

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

• A. $a^2$
• B. $b^2$
• C. $c^2$
• D. None of the above

Answer 1

Answer: (C)

$c^2$

In $\Delta$ ABC,

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

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

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

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

$=c^2$ ....... $[\because \cos c=\frac{a^2+b^2-c^2}{2ab}]$