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If tan⁻¹ y = tan⁻¹ x + tan⁻¹ ((2x)/(1-x²)), with |x| < (1)/(√(3)), then one possible value of y is:

1. (3x-x³)/(1+3x²)
2. (3x+x³)/(1+3x²)
3. (3x-x³)/(1-3x²)
4. (3x+x³)/(1-3x²)

Correct answer: (3x-x³)/(1-3x²)

Solution

For |x|<1, tan-1(2x/(1-x^2)) = 2 tan-1 x, so the RHS = 3 tan-1 x = tan-1((3x-x^3)/(1-3x^2)). Hence y = (3x-x^3)/(1-3x^2).

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