Trigonometry & Inverse Trigonometry Question - None

Question

Let tan⁻¹ y = tan⁻¹ x + tan⁻¹ \left(\frac{2x}{1-x^2}\right), where |x| < \frac{1}{\sqrt{3}}. Then, a value of y is

Accepted Answer

Step 1: Given, an^{-1} y = an^{-1} x + an^{-1} \left(\frac{2x}{1-x^2} ight), where |x| < \frac{1}{\sqrt{3}}

Step 2: Note that \frac{2x}{1-x^2} = \frac{2 an^{-1} x}{1-( an^{-1} x)^2}, which is the formula for an(2 an^{-1} x).

Step 3: Therefore, an^{-1}\left(\frac{2x}{1-x^2} ight) = 2 an^{-1} x

Step 4: So, an^{-1} y = an^{-1} x + 2 an^{-1} x = 3 an^{-1} x

Step 5: Thus, y = an(3 an^{-1} x) = \frac{3x - x^3}{1 - 3x^2} using the triple angle formula for tangent.

∴ Answer is (b).

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