Question

If \[ A+B+C=\pi, \] and \[ \cos B=\cos A \cos C, \] then \[ \tan A \tan C= \]

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(\dfrac{1}{2}\)

Hint:

Whenever angles satisfy \[ A+B+C=\pi, \] rewrite one angle in terms of the other two and use trigonometric sum identities for simplification.

Answer 1

The Correct Option is C

Step 1: Use the relation \(A+B+C=\pi\).
Given, \[ A+B+C=\pi \] Therefore, \[ B=\pi-(A+C) \] Taking cosine on both sides, \[ \cos B=\cos(\pi-(A+C)) \] Using the identity \[ \cos(\pi-\theta)=-\cos\theta, \] we get \[ \cos B=-\cos(A+C) \] Now expand: \[ \cos(A+C)=\cos A \cos C-\sin A \sin C \] Hence, \[ \cos B = -(\cos A \cos C-\sin A \sin C) \] \[ \cos B = -\cos A \cos C+\sin A \sin C \]

Step 2: Use the given condition.
Given, \[ \cos B=\cos A \cos C \] Substituting, \[ \cos A \cos C = -\cos A \cos C+\sin A \sin C \] Bring terms together: \[ 2\cos A \cos C=\sin A \sin C \]

Step 3: Divide by \(\cos A \cos C\).
Assuming \[ \cos A \neq0 \quad \text{and} \quad \cos C \neq0, \] divide both sides by \[ \cos A \cos C \] Then, \[ 2 = \frac{\sin A}{\cos A} \cdot \frac{\sin C}{\cos C} \] Thus, \[ 2=\tan A \tan C \] Therefore, \[ \tan A \tan C=2 \]

Step 4: Final conclusion.
Hence, \[ \boxed{2} \] which corresponds to option (3).