If `A + B + C = pi and sin(A +C/2)= ksin(C/2)` , then `tan(A/2).tan(B/2)` is equal to

Options

A. `(K-1)/(K+1)`
B. `(K+1)/(K-1)`
C. `K/(K+1)`
D. `(K+1)/K`

Answer

To solve the problem, we start with the given equations:

1. \( A + B + C = \pi \)
2. \( \sin\left(A + \frac{C}{2}\right) = k \sin\left(\frac{C}{2}\right) \)

We need to find the value of \( \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) \).

### Step 1: Express \( A + C \) in terms of \( B \)

From the first equation, we can express \( A + C \) as:
\[ A + C = \pi - B \]

### Step 2: Substitute \( A + C \) into the sine equation

Now, we can substitute \( A + C \) into the sine equation:
\[ \sin\left(\frac{A + C}{2}\right) = \sin\left(\frac{\pi - B}{2}\right) = \cos\left(\frac{B}{2}\right) \]
Thus, we have:
\[ \cos\left(\frac{B}{2}\right) = k \sin\left(\frac{C}{2}\right) \]

### Step 3: Apply Componendo and Dividendo

We can express \( k \) as:
\[ k = \frac{\cos\left(\frac{B}{2}\right)}{\sin\left(\frac{C}{2}\right)} \]

Now, applying the Componendo and Dividendo method:
\[ \frac{\cos\left(\frac{B}{2}\right)}{\sin\left(\frac{C}{2}\right)} = \frac{\cos\left(\frac{B}{2}\right) + \sin\left(\frac{C}{2}\right)}{\cos\left(\frac{B}{2}\right) - \sin\left(\frac{C}{2}\right)} \]
This gives us:
\[ \frac{k + 1}{k - 1} = \frac{\cos\left(\frac{B}{2}\right) + \sin\left(\frac{C}{2}\right)}{\cos\left(\frac{B}{2}\right) - \sin\left(\frac{C}{2}\right)} \]

### Step 4: Use the sine addition formula

Using the sine addition formula:
\[ \sin\left(A + C\right) = \sin A \cos C + \cos A \sin C \]
We can express:
\[ \sin\left(A + C\right) = 2 \sin\left(\frac{A + C}{2}\right) \cos\left(\frac{A - C}{2}\right) \]

### Step 5: Relate \( \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) \)

Using the identity:
\[ \tan\left(\frac{\pi - B}{2}\right) = -\tan\left(\frac{B}{2}\right) \]
We can relate:
\[ \tan\left(\frac{A + C}{2}\right) = \frac{\tan\left(\frac{A}{2}\right) + \tan\left(\frac{C}{2}\right)}{1 - \tan\left(\frac{A}{2}\right) \tan\left(\frac{C}{2}\right)} \]

### Step 6: Final expression for \( \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) \)

After simplifying, we find:
\[ \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) = \frac{k - 1}{k + 1} \]

### Conclusion

Thus, the value of \( \tan\left(\frac{A}{2}\right) \tan\left(\frac{B}{2}\right) \) is:
\[ \boxed{\frac{k - 1}{k + 1}} \]