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Question

Mathematics Question on Inverse Trigonometric Functions

Let tan−1y=tan−1x+tan−1(2x1−x2)tan^{-1}y=tan^{-1}x+tan^{-1}(\frac{2x}{1-x^2}). Then y is:

A

3x−x31−3x2\frac{3x-x^3}{1-3x^2}

B

3x+x31−3x2\frac{3x+x^3}{1-3x^2}

C

3x−x31+3x2\frac{3x-x^3}{1+3x^2}

D

3x+x31+3x2\frac{3x+x^3}{1+3x^2}

Answer

3x−x31−3x2\frac{3x-x^3}{1-3x^2}

Explanation

Solution

The correct option is (A):3x−x31−3x2\frac{3x-x^3}{1-3x^2}