Question

In a triangle ABC, if \(a - 2b + c = 0\), then \[ \cot \frac{A}{2} \cdot \cot \frac{C}{2} = ? \]

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Use half-angle formulas and multiply carefully; apply given linear relations between sides to simplify.

Answer 1

The Correct Option is C

Step 1: Recall formula for \(\cot \frac{A}{2}\).
\[ \cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}, \quad s = \frac{a+b+c}{2} \]

Step 2: Similarly for \(\cot \frac{C}{2}\).
\[ \cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}} \]

Step 3: Multiply the two.
\[ \cot \frac{A}{2} \cdot \cot \frac{C}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \frac{s}{s-b} \]

Step 4: Express \(s\) in terms of \(a, b, c\).
\[ s = \frac{a+b+c}{2} \Rightarrow s-b = \frac{a-b+c}{2} \]

Step 5: Apply given condition \(a - 2b + c = 0 \Rightarrow a+c = 2b\).
\[ s-b = \frac{(a+c)-b}{2} = \frac{2b - b}{2} = \frac{b}{2}, \quad s = \frac{a+b+c}{2} = \frac{(a+c)+b}{2} = \frac{2b + b}{2} = \frac{3b}{2} \]

Step 6: Compute final value.
\[ \cot \frac{A}{2} \cdot \cot \frac{C}{2} = \frac{s}{s-b} = \frac{3b/2}{b/2} = 3 \]
\[ \boxed{3} \]