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

Solution :
$$\eqalign{ & {\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right) \cr & \,\,\,\,\, = {\tan ^{ - 1}}x + 2{\tan ^{ - 1}}x = 3{\tan ^{ - 1}}x \cr & {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left[ {\frac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right] \cr & \Rightarrow \,\,y = \frac{{3x - {x^3}}}{{1 - 3{x^2}}} \cr} $$