Question

If the sides $a,b,c$ of $△ABC$ are in $A.P.$, prove that
$a{\cos }^2\frac{C}{2}+c{\cos }^2\frac{A}{2}=\frac{3b}{2}$

Answer 1

$L.H.S=\frac{1}{2}\left\{a(1+\cos C)\right\}+c\{1+\cos A\}$

$=\frac{1}{2}(a+c)+\frac{1}{2}\left\{a\cos C+ca\cos A\right\}$

$=b+\frac{1}{2}b=\frac{3}{2}b$.