Question

Let \(tan^{-1}y=tan^{-1}x+tan^{-1}(\frac{2x}{1-x^2})\). Then y is:

• A. \(\frac{3x-x^3}{1-3x^2}\)
• B. \(\frac{3x+x^3}{1-3x^2}\)
• C. \(\frac{3x-x^3}{1+3x^2}\)
• D. \(\frac{3x+x^3}{1+3x^2}\)

Answer 1

The Correct Option is A

Let \(y = \tan^{-1}x + \tan^{-1}\left(\frac{2x}{1-x^2}\right)\). According to the tangent addition formula: \[\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)\] Given the values \(a = x\) and \(b = \frac{2x}{1-x^2}\), we have: \[ \tan^{-1}y = \tan^{-1}\left(\frac{x+\frac{2x}{1-x^2}}{1-x\cdot\frac{2x}{1-x^2}}\right) \] Simplifying the expression, we have: \[y = \frac{x + \frac{2x}{1-x^2}}{1 - \frac{2x^2}{1-x^2}} \] Simplifying the numerator: \[x + \frac{2x}{1-x^2} = \frac{x(1-x^2) + 2x}{1-x^2} = \frac{x-x^3+2x}{1-x^2} = \frac{3x-x^3}{1-x^2}\] Simplifying the denominator: \[1 - \frac{2x^2}{1-x^2} = \frac{1-x^2-2x^2}{1-x^2} = \frac{1-3x^2}{1-x^2}\] Thus, substituting back, we have: \[y = \frac{\frac{3x-x^3}{1-x^2}}{\frac{1-3x^2}{1-x^2}} = \frac{3x-x^3}{1-3x^2}\] Therefore, \(y=\frac{3x-x^3}{1-3x^2}\).