In a `DeltaA B C ,1/(1+tan^2(A/2))+1/(1+tan^2(B/2))+1/(1+tan^2(C/2))=k [1+sin(A/2) sin(B/2) sin(C/2)],` then the value of `k` is

Answer

To solve the problem, we need to find the value of \( k \) in the equation:

\[ \frac{1}{1 + \tan^2\left(\frac{A}{2}\right)} + \frac{1}{1 + \tan^2\left(\frac{B}{2}\right)} + \frac{1}{1 + \tan^2\left(\frac{C}{2}\right)} = k \left[ 1 + \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right) \right] \]

### Step 1: Simplify the left-hand side

Using the identity \( 1 + \tan^2\theta = \sec^2\theta \), we can rewrite the left-hand side:

\[ \frac{1}{1 + \tan^2\left(\frac{A}{2}\right)} = \cos^2\left(\frac{A}{2}\right) \]

Thus, we can rewrite the left-hand side as:

\[ \cos^2\left(\frac{A}{2}\right) + \cos^2\left(\frac{B}{2}\right) + \cos^2\left(\frac{C}{2}\right) \]

### Step 2: Use the identity for cosine squared

Recall that:

\[ \cos^2\theta = \frac{1 + \cos(2\theta)}{2} \]

Applying this identity, we have:

\[ \cos^2\left(\frac{A}{2}\right) = \frac{1 + \cos(A)}{2}, \quad \cos^2\left(\frac{B}{2}\right) = \frac{1 + \cos(B)}{2}, \quad \cos^2\left(\frac{C}{2}\right) = \frac{1 + \cos(C)}{2} \]

Therefore, the left-hand side becomes:

\[ \frac{1 + \cos(A)}{2} + \frac{1 + \cos(B)}{2} + \frac{1 + \cos(C)}{2} \]

### Step 3: Combine terms

Combining the terms gives us:

\[ \frac{3 + \cos(A) + \cos(B) + \cos(C)}{2} \]

### Step 4: Use the identity for the sum of cosines

In a triangle, we know that:

\[ \cos(A) + \cos(B) + \cos(C) = 1 + \frac{r}{R} \]

where \( r \) is the inradius and \( R \) is the circumradius. However, for our purpose, we can just keep it as \( \cos(A) + \cos(B) + \cos(C) \).

### Step 5: Substitute into the equation

Now substituting back into the equation, we have:

\[ \frac{3 + \cos(A) + \cos(B) + \cos(C)}{2} = k \left[ 1 + \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right) \right] \]

### Step 6: Evaluate the right-hand side

Using the identity \( \sin\left(\frac{A}{2}\right) \sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right) \) can be expressed in terms of the sides of the triangle, but we can also evaluate it directly.

### Step 7: Find the value of \( k \)

By equating both sides and simplifying, we find that:

\[ k = 2 \]

Thus, the value of \( k \) is:

\[ \boxed{2} \]