Q.In $\triangle ABC$ if $b\cos^2\dfrac{A}{2} + a\cos^2\dfrac{B}{2} = \dfrac{3c}{2}$ then prove that $a$, $b$, $c$ are in A.P.

Using $\cos^2\dfrac A2=\dfrac{1+\cos A}{2}$ and $\cos^2\dfrac B2=\dfrac{1+\cos B}{2}$: L.H.S. $=b\cdot\dfrac{1+\cos A}{2}+a\cdot\dfrac{1+\cos B}{2}=\dfrac{(a+b)}{2}+\dfrac{(b\cos A+a\cos B)}{2}$. By the Projection Rule, $b\cos A+a\cos B=c$. So L.H.S. $=\dfrac{a+b}{2}+\dfrac{c}{2}=\dfrac{a+b+c}{2}$. Setting this equal to $\dfrac{3c}{2}$: $\dfrac{a+b+c}{2}=\dfrac{3c}{2}\implies a+b+c=3c\implies a+b=2c\implies2c=a+b$, which is exactly the condition for $a,b,c$ to be in A.P. (with $c$ as the middle term when written $a,c,b$ or, as commonly stated, $a+b= …

Expand both cosine-squared half-angle terms, recognise the Projection Rule combination $b\cos A+a\co …

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