A) \[k\]
B) \[2k\]
C) \[\frac{k}{2}\]
D) None of these
Correct Answer: C
Solution :
\[b{{\cos }^{2}}\frac{C}{2}+c{{\cos }^{2}}\frac{B}{2}\] \[=\frac{b}{2}(1+\cos C)+\frac{c}{2}(1+\cos B)\] \[=\frac{b}{2}+\frac{c}{2}+\frac{1}{2}(b\cos C+c\cos B)\] \[=\frac{b}{2}+\frac{c}{2}+\frac{a}{2}\] \[(\because \,\,a=b\cos C+c\cos B)\] \[=\frac{a+b+c}{2}=\frac{k}{2}\]You need to login to perform this action.
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