Question

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

• A. $s$
• B. $2s$
• C. $s/2$
• D. $noneofthese$

Answer 1

The correct option is A $s$

In $△ABC$

As we know that

$\cos \frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}},$$\cos \frac{B}{2}=\sqrt{\frac{s(s-b)}{ac}},$$\cos \frac{C}{2}=\sqrt{\frac{s(s-c)}{ba}}$

$b{\cos }^2\frac{C}{2}+c{\cos }^2\frac{B}{2}=b\times \frac{s(s-c)}{ab}+c\times \frac{s(s-b)}{ac}=\frac{s}{a}(s-c+s-b)=\frac{s}{a}(2s-(b+c\left)\right)=\frac{s}{a}\times a=s$