Answer: Let θ = arccos(x). Then, we have cos(θ) = x and sin(θ) = √(1 - x^2) (since θ is in the first quadrant, sin(θ) is positive).
Using the tangent-half-angle identity, we have:
tan(θ/2) = sin(θ)/(1 + cos(θ)) = √(1 - x^2)/(1 + x)
Therefore, we can express 4 tan(arccos(x)) as:
4 tan(arccos(x)) = 4 tan(θ/2) = 4(√(1 - x^2)/(1 + x))