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given:
![\therefore \tan^{-1}\left ( x+1 \right )+\tan^{-1}\left ( x-1 \right )= \tan^{-1}\left [ \frac{x+1+x-1}{1-\left [ \left ( x+1 \right )\left ( x-1 \right ) \right ]} \right ]](https://entrancecorner.codecogs.com/gif.latex?%5Ctherefore%20%5Ctan%5E%7B-1%7D%5Cleft%20%28%20x+1%20%5Cright%20%29+%5Ctan%5E%7B-1%7D%5Cleft%20%28%20x-1%20%5Cright%20%29%3D%20%5Ctan%5E%7B-1%7D%5Cleft%20%5B%20%5Cfrac%7Bx+1+x-1%7D%7B1-%5Cleft%20%5B%20%5Cleft%20%28%20x+1%20%5Cright%20%29%5Cleft%20%28%20x-1%20%5Cright%20%29%20%5Cright%20%5D%7D%20%5Cright%20%5D)
![\Rightarrow \tan^{-1}\left [ \frac{2x}{1-\left ( x^{2}-1 \right )} \right ]](https://entrancecorner.codecogs.com/gif.latex?%5CRightarrow%20%5Ctan%5E%7B-1%7D%5Cleft%20%5B%20%5Cfrac%7B2x%7D%7B1-%5Cleft%20%28%20x%5E%7B2%7D-1%20%5Cright%20%29%7D%20%5Cright%20%5D)
![\Rightarrow \tan^{-1}\left [ \frac{2x}{2-x^{2}} \right ]](https://entrancecorner.codecogs.com/gif.latex?%5CRightarrow%20%5Ctan%5E%7B-1%7D%5Cleft%20%5B%20%5Cfrac%7B2x%7D%7B2-x%5E%7B2%7D%7D%20%5Cright%20%5D)
we know that
Now,
So only answer is