Mark the tick against the correct answer in the following:

To Find: The value of $\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$

Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

$\Rightarrow \tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\tan ^{-1}\left(\frac{(1+x)+(1-x)}{1-(1+x)(1-x)}\right)$

$=\tan ^{-1}\left(\frac{2}{1-\left(1-x^{2}\right)}\right)$

$=\tan ^{-1}\left(\frac{2}{x^{2}}\right)$

Here since $\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$

$\Rightarrow \tan ^{-1}\left(\frac{2}{x^{2}}\right)=\frac{\pi}{2}$

$\Rightarrow \tan ^{-1}\left(\frac{2}{x^{2}}\right)=\tan ^{-1}(\infty)\left(\because \tan \frac{\pi}{2}=\infty\right)$

$\Rightarrow \frac{2}{x^{2}}=\infty$

$\Rightarrow x^{2}=\frac{2}{\infty}$

$\Rightarrow x=0$