Using the triple-angle identity for tangent, express tan(3A) purely in terms of tan(A). Which of the following formulas correctly represents tan(3A)?

Introduction / Context:
This trigonometry question focuses on the triple-angle identity for the tangent function. The goal is to express tan(3A) entirely in terms of tan(A). This formula is useful in advanced trigonometric manipulations, solving equations, and sometimes in integration and complex number applications. Knowing the correct form and being able to derive or verify it is a sign of strong trigonometric understanding.

Given Data / Assumptions:

• A is a real angle for which tan(A) is defined.
• We need to express tan(3A) in terms of tan(A) only.
• Several candidate formulas are provided; exactly one is correct.

Concept / Approach:
A standard identity for tangent of a sum is tan(X + Y) = (tan X + tan Y) / (1 − tan X * tan Y), provided the denominator is not zero. To derive tan(3A), we can think of 3A as A + 2A and use the double-angle formula tan(2A) in terms of tan(A). Alternatively, one can recall the known triple-angle identity: tan(3A) = (3 tan A − tan^3 A) / (1 − 3 tan^2 A). Here we check that this remembered identity is consistent with the addition formulas, and then match it to the given options.

Step-by-Step Solution:
1) Recall the double-angle formula for tangent: tan(2A) = (2 tan A) / (1 − tan^2 A). 2) To get tan(3A), write 3A = 2A + A and apply the sum formula: tan(3A) = tan(2A + A) = (tan 2A + tan A) / (1 − tan 2A * tan A). 3) Substitute tan 2A = (2 tan A) / (1 − tan^2 A). 4) Let t = tan A for simplicity. Then tan 2A = (2t) / (1 − t^2). 5) Compute the numerator: tan 2A + tan A = (2t / (1 − t^2)) + t = (2t + t(1 − t^2)) / (1 − t^2) = (2t + t − t^3) / (1 − t^2) = (3t − t^3) / (1 − t^2). 6) Compute the denominator: 1 − tan 2A * tan A = 1 − (2t / (1 − t^2)) * t = 1 − (2t^2 / (1 − t^2)). 7) Write 1 as (1 − t^2) / (1 − t^2) and combine: 1 − 2t^2 / (1 − t^2) = [(1 − t^2) − 2t^2] / (1 − t^2) = (1 − 3t^2) / (1 − t^2). 8) Therefore tan(3A) = [ (3t − t^3) / (1 − t^2) ] / [ (1 − 3t^2) / (1 − t^2) ] = (3t − t^3) / (1 − 3t^2). 9) Replace t by tan A to obtain tan(3A) = (3 tan A − tan^3 A) / (1 − 3 tan^2 A).

Verification / Alternative check:
We can test the formula with a simple angle, such as A = 15 degrees. Then 3A = 45 degrees and tan(45) = 1. Using tan A = tan 15 degrees, we know tan 15 is approximately 0.268. Substitute t ≈ 0.268 into (3t − t^3) / (1 − 3t^2). Numerically, 3t − t^3 ≈ 0.804 − 0.019 = 0.785 and 1 − 3t^2 ≈ 1 − 0.216 = 0.784. Their ratio is approximately 1.001, close to 1 within rounding errors, which matches tan 45 degrees. This supports the correctness of the identity.

Why Other Options Are Wrong:
Options b, c, and d have incorrect sign patterns in the numerator or denominator, which do not follow from the derivation using sum and double-angle formulas. Option e is structurally different and corresponds to neither a known triple-angle formula nor the derived expression. Only option a, (3tanA − tan^3A) / (1 − 3tan^2A), matches the proven identity exactly.

Common Pitfalls:
Some learners attempt to memorize the triple-angle formula but confuse the signs or swap the positions of terms. Others may attempt a derivation but make algebraic mistakes when simplifying complex fractions. Working with a substitution t = tan A and simplifying step by step reduces the chance of error and clarifies how the final formula arises from the basic tangent addition and double-angle identities.

Final Answer:
The correct triple-angle identity for tangent is (3tanA - tan^3A) / (1 - 3tan^2A).