In ∆ABC, prove the following:
Question:

In ∆ABC, prove the following:

$4\left(b c \cos ^{2} \frac{A}{2}+c a \cos ^{2} \frac{B}{2}+a b \cos ^{2} \frac{C}{2}\right)=(a+b+c)^{2}$

Solution:

LHS

$=4\left(b c \cos ^{2} \frac{A}{2}+c a \cos ^{2} \frac{B}{2}+a b \cos ^{2} \frac{C}{2}\right)$

$=4\left[b c\left(\frac{1+\cos A}{2}\right)+c a\left(\frac{1+\cos B}{2}\right)+a b\left(\frac{1+\cos C}{2}\right)\right]$

$=2 b c+2 b c \cos A+2 c a+2 c a \cos B+2 a b+2 a b \cos C$

$=2(a b+b c+c a)+2 b c\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)+2 c a\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)+2 a b\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)$

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

$=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 a c$

$=(a+b+c)^{2}=\mathrm{RHS}$

Hence proved.

Administrator

Leave a comment

Please enter comment.
Please enter your name.