Question

If \( A+B+C=\pi \) and \( \cos A = \cos B \cos C \), then \( \tan A - \tan B - \tan C = \)

• A. \( -1 \)
• B. \( 0 \)
• C. \( 1 \)
• D. \( 2 \)

Hint:

When given \( \cos A = \cos B \cos C \) in a triangle, always use the expansion of \( \cos(B+C) \) to find the relationship \( \tan B \tan C = 2 \), which is a common shortcut for these problems.

Answer 1

The Correct Option is B

Step 1: Use the angle sum.
Since $A + B + C = \pi$, we have $A = \pi - (B + C)$, so $\cos A = -\cos(B+C)$.
Step 2: Apply the given condition.
The condition $\cos A = \cos B \cos C$ becomes $-\cos(B+C) = \cos B \cos C$.
Step 3: Expand $\cos(B+C)$.
Using $\cos(B+C) = \cos B\cos C - \sin B\sin C$: \[ -(\cos B\cos C - \sin B\sin C) = \cos B\cos C, \] so $\sin B\sin C = 2\cos B\cos C$.
Step 4: Get $\tan B\tan C$.
Divide by $\cos B\cos C$: $\tan B\tan C = 2$.
Step 5: Find $\tan A$.
Since $A = \pi - (B+C)$, $\tan A = -\tan(B+C) = -\dfrac{\tan B + \tan C}{1 - \tan B\tan C} = -\dfrac{\tan B + \tan C}{1 - 2} = \tan B + \tan C$.
Step 6: Evaluate the expression.
Then $\tan A - \tan B - \tan C = (\tan B + \tan C) - \tan B - \tan C = 0$, which is option (B).
\[ \boxed{0} \]