Rewrite 6 tan 3x in terms of tan x.
Actually, the answer is 6 * (3tan(x) - tan^3(x))/(1 - 3tan^2(x)), due to the triple angle formula for tangent. To derive it, you must use the angle sum formula for tangent twice:
6tan(3x) = 6tan(x + 2x) = 6 * (tan(x) + tan(2x))/(1 - tan(x)tan(2x)) = 6 * (tan(x) + [2tan(x)]/[1 - tan^2(x)])/(1 - tan(x) * [2tan(x)]/[1 - tan^2(x)]) = 6 * (tan(x) - tan^3(x) + 2tan(x))/(1 - tan^2(x) - tan(x) * 2tan(x)) = 6 * (3tan(x) - tan^3(x))/(1 - 3tan^2(x)).
You are correct. I apologize for the oversight in my initial response. The correct answer is indeed 6 * (3tan(x) - tan^3(x))/(1 - 3tan^2(x)). Thank you for providing the accurate derivation.
Let's use the half-angle formula for tangent, which states that tan(2θ) = 2tan(θ) / (1 - tan^2(θ)).
tan(2θ) = 2tan(θ) / (1 - tan^2(θ))
tan(3x) = 3tan(x) / (1 - tan^2(x))
Therefore, 6tan(3x) = 6(3tan(x) / (1 - tan^2(x)))
= 18tan(x) / (1 - tan^2(x))