Question

In \(\triangle ABC\), if \(3a = b + c\), then value of \(\cot\frac{B}{2}\cot\frac{C}{2}\) is:

• A. 1
• B. 2
• C. 3
• D. \(\frac{1}{2}\)

Hint:

Use \(\cot\frac{B}{2}\cot\frac{C}{2} = \frac{s}{s-a}\) and \(\cot\frac{A}{2} = \frac{s-a}{r}\).

Answer 1

The Correct Option is B

Concept: In any triangle, \(\cot\frac{B}{2} = \frac{s-b}{r}\) and \(\cot\frac{C}{2} = \frac{s-c}{r}\), where \(s\) is semi-perimeter and \(r\) is inradius.

Step 1: Given \(3a = b + c\). Semi-perimeter: \(s = \frac{a+b+c}{2} = \frac{a + (b+c)}{2} = \frac{a + 3a}{2} = \frac{4a}{2} = 2a\).

Step 2: Compute \(s-b\) and \(s-c\). \[ s - b = 2a - b,\quad s - c = 2a - c \] Also \(b + c = 3a \Rightarrow c = 3a - b\).

Step 3: Product of cotangents. \[ \cot\frac{B}{2}\cot\frac{C}{2} = \frac{(s-b)(s-c)}{r^2} \] Using the identity \((s-b)(s-c) = r^2 + (s-b)(s-c)\)? Alternatively, use standard formula: \[ \cot\frac{B}{2}\cot\frac{C}{2} = \frac{s(s-b)(s-c)}{\Delta^2} \cdot \ldots \] Simpler: Using \(\cot\frac{B}{2} = \frac{s(s-b)}{\Delta}\) etc., the product simplifies to: \[ \cot\frac{B}{2}\cot\frac{C}{2} = \frac{s}{s-a} \] Given \(s = 2a\) and \(s-a = 2a - a = a\): \[ \frac{s}{s-a} = \frac{2a}{a} = 2 \]