Question

In $\Delta ABC$, $a-b^2{\cos }^2\frac{c}{2}+a+b^2{\sin }^2\frac{c}{2}$ is equal to

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

Answer 1

Answer: (C)

$c^2$

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

$=a^2+b^2-(a^2+b^2-c^2)$
$=c^2$