If the sides a,b,c of `DeltaABC` are in A.p., then `cot ""1/2A,cot""1/2B, cot""1/2C` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. None

Answer

To prove that if the sides \( a, b, c \) of triangle \( \Delta ABC \) are in Arithmetic Progression (AP), then \( \cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2} \) are also in AP, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Condition of AP**:
Since \( a, b, c \) are in AP, we have:
\[ 2b = a + c \]
where \( a = BC, b = CA, c = AB \).

**Hint**: Remember that for three numbers to be in AP, the middle number must be the average of the other two.

2. **Using the Semi-Perimeter**:
Let \( s \) be the semi-perimeter of the triangle:
\[ s = \frac{a + b + c}{2} \]

3. **Expressing Cotangent in Terms of Sides**:
The cotangent of half-angles can be expressed as:
\[ \cot \frac{A}{2} = \frac{s(s-a)}{K}, \quad \cot \frac{B}{2} = \frac{s(s-b)}{K}, \quad \cot \frac{C}{2} = \frac{s(s-c)}{K} \]
where \( K \) is the area of the triangle.

**Hint**: Familiarize yourself with the relationship between the angles of a triangle and its sides.

4. **Setting Up the AP Condition**:
We need to show that:
\[ 2 \cot \frac{B}{2} = \cot \frac{A}{2} + \cot \frac{C}{2} \]

5. **Substituting the Cotangent Values**:
Substitute the expressions for \( \cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2} \):
\[ 2 \cdot \frac{s(s-b)}{K} = \frac{s(s-a)}{K} + \frac{s(s-c)}{K} \]

**Hint**: Notice that \( K \) cancels out since it is common in all terms.

6. **Simplifying the Equation**:
After canceling \( K \):
\[ 2s(s-b) = s(s-a) + s(s-c) \]

7. **Distributing and Rearranging**:
Expanding both sides gives:
\[ 2s^2 - 2bs = s^2 - as + s^2 - cs \]
Combine like terms:
\[ 2s^2 - 2bs = 2s^2 - (a + c)s \]

8. **Final Simplification**:
Rearranging leads to:
\[ -2bs = -(a + c)s \]
Dividing by \( -s \) (assuming \( s \neq 0 \)):
\[ 2b = a + c \]
This is the condition for \( a, b, c \) being in AP.

9. **Conclusion**:
Since we have shown that \( 2 \cot \frac{B}{2} = \cot \frac{A}{2} + \cot \frac{C}{2} \), we conclude that \( \cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2} \) are in AP.