In a triangle ABC , if cos A = cos B cos C , what is the value of tan A- tan B - tan C ?

Options

A. 0
B. `-1`
C. 1 + tan A tan B tan C
D. tan A tan B tan C- 1

Explanation

To solve the problem, we need to find the value of \( \tan A - \tan B - \tan C \) given that \( \cos A = \cos B \cos C \) in triangle ABC.

### Step-by-Step Solution:

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

2. **Express \( \tan A \), \( \tan B \), and \( \tan C \)**:
We know that:
\[ \tan A = \frac{\sin A}{\cos A}, \quad \tan B = \frac{\sin B}{\cos B}, \quad \tan C = \frac{\sin C}{\cos C} \]
Therefore, we can write:
\[ \tan A - \tan B - \tan C = \frac{\sin A}{\cos A} - \frac{\sin B}{\cos B} - \frac{\sin C}{\cos C} \]

3. **Combine the terms over a common denominator**:
The common denominator will be \( \cos A \cos B \cos C \):
\[ = \frac{\sin A \cos B \cos C - \sin B \cos A \cos C - \sin C \cos A \cos B}{\cos A \cos B \cos C} \]

4. **Substitute \( \cos A \) with \( \cos B \cos C \)**:
Since \( \cos A = \cos B \cos C \), we can substitute this in:
\[ = \frac{\sin A \cos B \cos C - \sin B (\cos B \cos C) \cos C - \sin C (\cos B \cos C) \cos B}{\cos A \cos B \cos C} \]

5. **Simplify the numerator**:
The numerator simplifies to:
\[ = \sin A \cos B \cos C - \sin B \cos^2 C - \sin C \cos^2 B \]

6. **Use the sine rule**:
In triangle ABC, we know that \( A + B + C = \pi \). Thus, we can use the identity \( \sin(B + C) = \sin(\pi - A) = \sin A \):
\[ \sin(B + C) = \sin A \]

7. **Apply the sine addition formula**:
\[ \sin(B + C) = \sin B \cos C + \cos B \sin C \]
Therefore, we have:
\[ \sin A = \sin B \cos C + \cos B \sin C \]

8. **Substituting back**:
Substitute this back into the expression:
\[ \tan A - \tan B - \tan C = \frac{\sin A - \sin(B + C)}{\cos A \cos B \cos C} \]
Since \( \sin A - \sin A = 0 \):
\[ = 0 \]

### Conclusion:
Thus, the value of \( \tan A - \tan B - \tan C \) is:
\[ \boxed{0} \]