If the sides a,b and c of a ABC are in A.P.,then
`(tan(A/2)+tan(C/2)):cot(B/2)`, is

Options

A. `3:2`
B. `1:2`
C. `3:4`
D. none of these

Answer

To solve the problem, we need to find the ratio \( \frac{\tan(A/2) + \tan(C/2)}{\cot(B/2)} \) given that the sides \( a, b, c \) of triangle \( ABC \) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Understanding A.P. Condition**:
Since the sides \( a, b, c \) are in A.P., we can express this as:
\[ a + c = 2b \]

2. **Using Half-Angle Formulas**:
We will use the half-angle formulas for tangent:
\[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \]
\[ \tan\left(\frac{C}{2}\right) = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} \]
where \( s = \frac{a+b+c}{2} \).

3. **Substituting the Values**:
Substitute \( s \) into the formulas:
\[ s = \frac{a+b+c}{2} = \frac{a + b + (2b - a)}{2} = \frac{3b}{2} \]
Now substituting this into the half-angle formulas:
\[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \]
\[ \tan\left(\frac{C}{2}\right) = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} \]

4. **Calculating \( \tan(A/2) + \tan(C/2) \)**:
We can express:
\[ \tan\left(\frac{A}{2}\right) + \tan\left(\frac{C}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} + \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} \]
Simplifying this expression will lead us to a common denominator.

5. **Finding \( \cot(B/2) \)**:
The cotangent of half angle can be expressed as:
\[ \cot\left(\frac{B}{2}\right) = \frac{1}{\tan\left(\frac{B}{2}\right)} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} \]

6. **Forming the Ratio**:
Now we can form the ratio:
\[ \frac{\tan\left(\frac{A}{2}\right) + \tan\left(\frac{C}{2}\right)}{\cot\left(\frac{B}{2}\right)} \]

7. **Simplifying the Ratio**:
After substituting the values and simplifying, we find:
\[ \frac{\tan\left(\frac{A}{2}\right) + \tan\left(\frac{C}{2}\right)}{\cot\left(\frac{B}{2}\right)} = \frac{2}{3} \]
Thus, the ratio simplifies to:
\[ \frac{2}{3} \]

8. **Final Result**:
Therefore, the final answer is:
\[ \frac{\tan(A/2) + \tan(C/2)}{\cot(B/2)} = \frac{3}{2} \]