Question

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

• A. $\lambda$
• B. $\frac{\lambda }{2}$
• C. $2\lambda$
• D. none of these.

Answer 1

Answer: (B)

$\frac{\lambda }{2}$

$a+b+c=\lambda =2s$
$\therefore \lambda =2s$ ................$\left(1\right)$
$\Rightarrow s=\frac{\lambda }{2}$
$b{\cos }^2\left(\frac{C}{2}\right)+c{\cos }^2\left(\frac{B}{2}\right)$
$=b\frac{s(s-c)}{ab}+c\frac{s(s-c)}{ac}$
On simplification we get
$=\frac{s}{a}[s-c+s-b]$
$=\frac{s}{a}(2s-\{b+c\})$
$=\frac{s}{a}(2s-\{2s-a\})$
$=\frac{s}{a}\times a$
$\Rightarrow s=\frac{\lambda }{2}$ from $\left(1\right)$