In a triangle ABC, \(a\cos ^2\frac{C}{2}+\, c\, \, {\cos }^2\frac{A}{2}=\frac{3b}{2}\) then the sides of the triangle are in
Step-by-step Solution:
Using the half angle formula: \[ a \left( \frac{s(s - c)}{ab} \right) + c \left( \frac{s(s - a)}{bc} \right) = \frac{3b}{2} \] \[ \Rightarrow s(s - c) + s(s - a) = \frac{3b^2}{2} \] \[ \Rightarrow s(2s - a - c) = \frac{3b^2}{2} \] Or \[ \frac{b(a + b + c)}{2} = \frac{3b}{2} \Rightarrow a + c = 2b \] The sides are in AP.