Question

In a triangle \(A B C\), with usual notations, \(3 \mathrm{~b}=\mathrm{a}+\mathrm{c}\), then \(\cot \frac{\mathrm{A}}{2} \cdot \cot \frac{\mathrm{C}}{2}=\)

• A. 1
• B. 2
• C. \(\frac{1}{2}\)
• D. 4

Hint:

Use the half-angle cotangent formulas in terms of s and the sides.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In a triangle, \(\cot\frac A2=\frac{s-a}{r}\) and \(\cot\frac C2=\frac{s-c}{r}\).

Step 2: Product
\(\cot\frac A2\cot\frac C2=\frac{(s-a)(s-c)}{r^2}=\frac{(s-a)(s-c)\,s}{(s-a)(s-b)(s-c)}=\frac{s}{s-b}\).

Step 3: Use 3b = a + c
\(a+b+c=4b\), so \(s=2b\) and \(s-b=b\).

Step 4: Result
\(\frac{s}{s-b}=\frac{2b}{b}=2\).

Final Answer:
The product equals 2. \[ \boxed{\text{(B) }2} \]