If `A+B+C=pi, and cosA=cosB*cosC` then `tanB*tanC` has the value equal to

Options

A. 1
B. `1/2`
C. 2
D. 3

Answer

To solve the problem where \( A + B + C = \pi \) and \( \cos A = \cos B \cdot \cos C \), we need to find the value of \( \tan B \cdot \tan C \).

### Step-by-Step Solution:

1. **Express \( A \) in terms of \( B \) and \( C \)**:
\[ A = \pi - (B + C) \]

**Hint**: Use the fact that the sum of angles in a triangle is \( \pi \).

2. **Use the cosine identity**:
\[ \cos A = \cos(\pi - (B + C)) = -\cos(B + C) \]

**Hint**: Recall that \( \cos(\pi - x) = -\cos x \).

3. **Apply the cosine addition formula**:
\[ -\cos(B + C) = -(\cos B \cos C - \sin B \sin C) \]
Thus,
\[ \cos A = \cos B \cos C \]
leads to:
\[ -(\cos B \cos C - \sin B \sin C) = \cos B \cos C \]

**Hint**: Use the identity \( \cos(A + B) = \cos A \cos B - \sin A \sin B \).

4. **Rearranging the equation**:
\[ -\cos B \cos C + \sin B \sin C = \cos B \cos C \]
This simplifies to:
\[ \sin B \sin C = 2\cos B \cos C \]

**Hint**: Combine like terms and isolate the sine and cosine products.

5. **Divide both sides by \( \cos B \cos C \)**:
\[ \frac{\sin B \sin C}{\cos B \cos C} = 2 \]
This can be rewritten using the definition of tangent:
\[ \tan B \tan C = 2 \]

**Hint**: Remember that \( \tan x = \frac{\sin x}{\cos x} \).

6. **Conclusion**:
Therefore, the value of \( \tan B \cdot \tan C \) is:
\[ \tan B \tan C = 2 \]

### Final Answer:
The value of \( \tan B \cdot \tan C \) is \( 2 \).