Let tan⁻¹(y) = tan⁻¹(x) + tan⁻¹(2x/(1 - x²)), where |x|

Question

Let tan⁻¹(y) = tan⁻¹(x) + tan⁻¹(2x/(1 - x²)), where |x| < 1/sqrt(3). Then a value of y is:

Accepted Answer

(3x - x³)/(1 - 3x²). For |x| < 1, tan⁻¹(2x/(1 - x²)) = 2 tan⁻¹(x). Hence tan⁻¹(y) = tan⁻¹(x) + 2 tan⁻¹(x) = 3 tan⁻¹(x). Let theta = tan⁻¹(x), so tan(theta) = x. Then y = tan(3 theta) = (3 tan theta - tan³ theta)/(1 - 3 tan² theta) = (3x - x³)/(1 - 3x²). The condition |x| < 1/sqrt(3) keeps 1 - 3x² > 0, valid.

Let tan⁻¹(y) = tan⁻¹(x) + tan⁻¹(2x/(1 - x²)), where |x| < 1/sqrt(3). Then a value of y is:

Solution

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