In ∆ABC, prove the following:

RHS $=(b+c)^{2}-4 b c \cos ^{2} \frac{A}{2}$

$=b^{2}+c^{2}+2 b c-4 b c\left(\frac{1+\cos A}{2}\right)$

$=b^{2}+c^{2}+2 b c-2 b c(1+\cos A)$

$=b^{2}+c^{2}+2 b c(1-1-\cos A)$

$=b^{2}+c^{2}-2 b c \cos A$

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

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

$=a^{2}=\mathrm{LHS}$

Hence proved.