case 1
\[ tan^{-1}(\frac{2x}{1-x^2})=2tan^{-1}(x)\]
LHS=\[ tan^{-1}(\frac{2x}{1-x^2})\]
let x= \[tan(\theta )\] \[ \theta =tan^{-1}(x)\]
\[ \theta \in (\frac{-\Pi }{2},\frac{\Pi }{2})\]
Then LHS=\[tan^{-1}( \frac{2tan\theta}{1-tan^2\theta } )=tan^{-1}(tan2\theta )=2\theta \] {since \[ tan^{-1}(tanx)=x\]
=\[ 2tan^{-1}(x)\]
Boundary conditions for case 1
\[-1 < x < 1\]
case 2
For \[\Pi +\]\[ tan^{-1}(\frac{2x}{1-x^2})=2tan^{-1}(x)\]
boundary condition is x > 1
Case 3
For \[-\Pi +\]\[ tan^{-1}(\frac{2x}{1-x^2})=2tan^{-1}(x)\]
Boundary condition for case 3 is x < -1
