Question

If $k$ be the perimeter of the $△ABC$, then $b{\cos }^2\frac{C}{2}+c{\cos }^2\frac{B}{2}$ is equal to

• A. $k$
• B. $2k$
• C. $\frac{k}{2}$
• D. None of these

Answer 1

Answer: (C)

$\frac{k}{2}$

The value of $b{\cos }^2\frac{C}{2}+c{\cos }^2\frac{B}{2}$ is
$=\frac{b}{2}(1+\cos C)+\frac{c}{2}(1+\cos B)$
$=\frac{b}{2}+\frac{c}{2}+\frac{1}{2}(b\cos C+c\cos B)$
$=\frac{b}{2}+\frac{c}{2}+\frac{a}{2}(\because a=b\cos C+c\cos B)$
$=\frac{a+b+c}{2}=\frac{k}{2}$.