in any triangle ABC, which is not right angled `Sigma cosA.cosecB.cosecC` is equal to

Options

A. 1
B. 2
C. 3
D. None of these

Answer

To solve the problem, we need to find the value of the expression \( \Sigma \cos A \cdot \csc B \cdot \csc C \) in triangle ABC, which is not a right-angled triangle.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We need to evaluate \( \Sigma \cos A \cdot \csc B \cdot \csc C \), which can be rewritten as:
\[ \cos A \cdot \csc B \cdot \csc C + \cos B \cdot \csc A \cdot \csc C + \cos C \cdot \csc A \cdot \csc B \]

2. **Using the Definition of Cosecant**:
Recall that \( \csc B = \frac{1}{\sin B} \) and \( \csc C = \frac{1}{\sin C} \). Thus, we can rewrite the expression:
\[ = \cos A \cdot \frac{1}{\sin B} \cdot \frac{1}{\sin C} + \cos B \cdot \frac{1}{\sin A} \cdot \frac{1}{\sin C} + \cos C \cdot \frac{1}{\sin A} \cdot \frac{1}{\sin B} \]

3. **Finding a Common Denominator**:
The common denominator for the three terms is \( \sin A \sin B \sin C \). Thus, we can write:
\[ = \frac{\cos A \cdot \sin A + \cos B \cdot \sin B + \cos C \cdot \sin C}{\sin A \sin B \sin C} \]

4. **Using the Identity**:
We can use the identity \( \sin 2A = 2 \sin A \cos A \), \( \sin 2B = 2 \sin B \cos B \), and \( \sin 2C = 2 \sin C \cos C \):
\[ = \frac{\frac{1}{2} \sin 2A + \frac{1}{2} \sin 2B + \frac{1}{2} \sin 2C}{\sin A \sin B \sin C} \]

5. **Applying the Known Identity**:
We know from trigonometric identities that:
\[ \sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C \]
Therefore, substituting this into our expression gives:
\[ = \frac{\frac{1}{2} \cdot 4 \sin A \sin B \sin C}{\sin A \sin B \sin C} = 2 \]

6. **Conclusion**:
Thus, the value of \( \Sigma \cos A \cdot \csc B \cdot \csc C \) in any triangle ABC that is not right-angled is:
\[ \boxed{2} \]