Question

With usual notations, in \( \triangle ABC \), if \( b\cos^2\frac{C}{2} + c\cos^2\frac{B}{2} = \frac{3a}{2} \), then

• A. \( b, a, c \) are in A.P.
• B. \( b, a, c \) are in G.P.
• C. \( a, b, c \) are in G.P.
• D. \( a, b, c \) are in A.P.

Hint:

Half-angle identities often help convert trigonometric conditions into simple algebraic relations among sides.

Answer 1

The Correct Option is A

Step 1: Use half-angle identities.
\[ \cos^2\frac{B}{2} = \frac{s(s-b)}{ac}, \qquad \cos^2\frac{C}{2} = \frac{s(s-c)}{ab} \] where \( s = \frac{a+b+c}{2} \).

Step 2: Substitute in the given expression.
\[ b\cdot\frac{s(s-c)}{ab} + c\cdot\frac{s(s-b)}{ac} = \frac{3a}{2} \] \[ \frac{s}{a}[(s-c)+(s-b)] = \frac{3a}{2} \]

Step 3: Simplify.
\[ \frac{s}{a}(2s-b-c) = \frac{3a}{2} \] But \( 2s = a+b+c \), hence \[ \frac{s}{a}\,a = \frac{3a}{2} \Rightarrow s = \frac{3a}{2} \]

Step 4: Conclude the relation among sides.
\[ \frac{a+b+c}{2} = \frac{3a}{2} \Rightarrow b+c = 2a \] Thus, \( b, a, c \) are in arithmetic progression.