Formula to evaluate tan inverse 2x minus one minus x square

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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