Question

$\tan 3A \cdot \tan 2A \cdot \tan A =$

• A. $\tan 3A + \tan 2A - \tan A$
• B. $\tan 3A - \tan 2A - \tan A$
• C. $\tan 3A + \tan 2A + \tan A$
• D. $\tan 3A - \tan 2A + \tan A$

Hint:

Whenever three angles satisfy $X = Y + Z$, their tangents satisfy the universal product-to-difference conversion property: $$\tan X \cdot \tan Y \cdot \tan Z = \tan X - \tan Y - \tan Z$$ Recognizing this relationship structure lets you skip the derivation entirely.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem asks for an alternate identity format for a product of tangent functions whose angle arguments are related linearly ($3A = 2A + A$).

Step 2: Key Formula or Approach:
We utilize the compound angle expansion formula for tangent: $$\tan(X + Y) = \frac{\tan X + \tan Y}{1 - \tan X \tan Y}$$ Setting $X = 2A$ and $Y = A$ yields an algebraic relation that connects the product and differences.

Step 3: Detailed Explanation:
Write the angle relation explicitly: $$3A = 2A + A$$ Taking tangent on both sides: $$\tan 3A = \tan(2A + A)$$ Apply the tangent addition formula to the right-hand side: $$\tan 3A = \frac{\tan 2A + \tan A}{1 - \tan 2A \tan A}$$ Cross-multiplying to clear the fraction: $$\tan 3A (1 - \tan 2A \tan A) = \tan 2A + \tan A$$ $$\tan 3A - \tan 3A \cdot \tan 2A \cdot \tan A = \tan 2A + \tan A$$ Rearranging terms to isolate the product expression $\tan 3A \cdot \tan 2A \cdot \tan A$: $$\tan 3A - \tan 2A - \tan A = \tan 3A \cdot \tan 2A \cdot \tan A$$ $$\Rightarrow \tan 3A \cdot \tan 2A \cdot \tan A = \tan 3A - \tan 2A - \tan A$$

Step 4: Final Answer:
The calculated equation matches exactly with option (B).