In a triangle ABC, if
`cot A + cot B + cotC =cot theta `, then
`( sin A sin B sin C )/( 1+ cos A cos B cos C ) =`

Options

A. `tan^(2) theta`
B. ` cot ^(2) theta`
C. `cot theta`
D. `tan theta `

Answer

To solve the problem, we need to find the value of \(\frac{\sin A \sin B \sin C}{1 + \cos A \cos B \cos C}\) given that \(\cot A + \cot B + \cot C = \cot \theta\).

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ \cot A + \cot B + \cot C = \cot \theta \]

2. **Use the identity for cotangent**:
Recall that \(\cot A = \frac{\cos A}{\sin A}\). Therefore, we can rewrite the equation as:
\[ \frac{\cos A}{\sin A} + \frac{\cos B}{\sin B} + \frac{\cos C}{\sin C} = \cot \theta \]

3. **Combine the cotangent terms**:
The left-hand side can be combined into a single fraction:
\[ \frac{\cos A \sin B \sin C + \cos B \sin A \sin C + \cos C \sin A \sin B}{\sin A \sin B \sin C} = \cot \theta \]

4. **Cross-multiply**:
This gives us:
\[ \cos A \sin B \sin C + \cos B \sin A \sin C + \cos C \sin A \sin B = \cot \theta \cdot \sin A \sin B \sin C \]

5. **Rearranging the equation**:
We can express the left-hand side as:
\[ 1 + \cos A \cos B \cos C = \frac{\sin A \sin B \sin C}{\cot \theta} \]

6. **Substituting back into the original expression**:
Now we need to find:
\[ \frac{\sin A \sin B \sin C}{1 + \cos A \cos B \cos C} \]

7. **Using the derived equation**:
Substitute the expression we derived:
\[ \frac{\sin A \sin B \sin C}{\cot \theta \cdot \sin A \sin B \sin C} = \frac{1}{\cot \theta} \]

8. **Final simplification**:
Since \(\frac{1}{\cot \theta} = \tan \theta\), we conclude that:
\[ \frac{\sin A \sin B \sin C}{1 + \cos A \cos B \cos C} = \tan \theta \]

### Final Answer:
\[ \frac{\sin A \sin B \sin C}{1 + \cos A \cos B \cos C} = \tan \theta \]