Mark the tick against the correct answer in the following:
If $\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$ then $x=?$
A. 1
B. $-1$
C. 0
D. $\frac{1}{2}$
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 $\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$
$\Rightarrow \tan ^{-1}\left(\frac{2}{x^{2}}\right)=\frac{\pi}{2}$
$\Rightarrow \tan ^{-1}\left(\frac{2}{x^{2}}\right)=\tan ^{-1}(\infty)\left(\because \tan \frac{\pi}{2}=\infty\right)$
$\Rightarrow \frac{2}{x^{2}}=\infty$
$\Rightarrow x^{2}=\frac{2}{\infty}$
$\Rightarrow x=0$
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