In any triangle ABC, if sin A , sin B, sin C are in AP, then the maximum value of `tan ""B/2` is

Options

A. `-(1)/(sqrt3)`
B. `(1)/(sqrt3)`
C. `1/3`
D. None of these

Answer

To find the maximum value of \( \tan \frac{B}{2} \) given that \( \sin A, \sin B, \sin C \) are in arithmetic progression (AP), we can follow these steps:

### Step 1: Understanding the Condition
Since \( \sin A, \sin B, \sin C \) are in AP, we can express this condition mathematically as:
\[ 2 \sin B = \sin A + \sin C \]

### Step 2: Using the Sine Rule
In triangle \( ABC \), we know that:
\[ \sin C = \sin(A + B) = \sin A \cos B + \cos A \sin B \]
Substituting this into our AP condition gives:
\[ 2 \sin B = \sin A + \sin A \cos B + \cos A \sin B \]

### Step 3: Rearranging the Equation
Rearranging the equation, we get:
\[ 2 \sin B - \cos A \sin B = \sin A + \sin A \cos B \]
Factoring out \( \sin B \) on the left side:
\[ \sin B (2 - \cos A) = \sin A (1 + \cos B) \]

### Step 4: Expressing \( \tan \frac{B}{2} \)
Using the half-angle formula, we know:
\[ \tan \frac{B}{2} = \frac{\sin B}{1 + \cos B} \]
Thus, we can express \( \tan \frac{B}{2} \) in terms of \( \sin A \) and \( \cos A \):
\[ \tan \frac{B}{2} = \frac{\sin A (1 + \cos B)}{(2 - \cos A)(1 + \cos B)} \]

### Step 5: Finding the Maximum Value
To find the maximum value of \( \tan \frac{B}{2} \), we need to analyze the expression:
\[ \tan \frac{B}{2} = \frac{\sin A}{2 - \cos A} \]
We can use the identity \( \sin^2 A + \cos^2 A = 1 \) to express \( \sin A \) in terms of \( \cos A \):
Let \( \cos A = x \), then \( \sin A = \sqrt{1 - x^2} \).

Substituting this into our expression gives:
\[ \tan \frac{B}{2} = \frac{\sqrt{1 - x^2}}{2 - x} \]

### Step 6: Maximizing the Function
To find the maximum value, we can differentiate \( \tan \frac{B}{2} \) with respect to \( x \) and set the derivative to zero. However, we can also analyze the boundaries of \( x \) (which ranges from -1 to 1) to find the maximum value.

After performing the necessary calculations, we find that:
\[ \max \tan \frac{B}{2} = \frac{1}{\sqrt{3}} \]

### Conclusion
Thus, the maximum value of \( \tan \frac{B}{2} \) is:
\[ \boxed{\frac{1}{\sqrt{3}}} \]