MHTCET 16 Apr 2026 Shift 1 Paper

Concept:Use the half-angle identity $\cos^2 \frac{\theta}{2} = \frac{1+\cos \theta}{2}$ and the cosine rule to simplify the given condition.Explanation:Rewrite each term using the half-angle formula:$$b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = \frac{b(1+\cos C) + c(1+\cos B)}{2}.$$The given condition becomes:$$\frac{b+c + b\cos C + c\cos B}{2} = \frac{3a}{2}.$$From the cosine rule, we have the standard identity:$$b\cos C + c\cos B = a.$$Substitute this into the equation:$$\frac{b+c+a}{2} = \frac{3a}{2}.$$Multiply both sides by $2$:$$b+c+a = 3a \Rightarrow b+c = 2a.$$Thus, $a$ is the arithmetic mean of $b$ and $c$, so $b, a, c$ are in A.P.Answer:Option C: $b, a, c$ are in A.P.