A) \[x=1\]
B) \[x=-1\]
C) \[x=0\]
D) \[x=\frac{1}{2}\]
Correct Answer: C
Solution :
[c] \[{{\tan }^{-1}}(1+x)+ta{{n}^{-1}}(1-x)=\frac{\pi }{2}\] \[{{\tan }^{-1}}\left[ \frac{(1+x)+(1-x)}{1-(1+x)(1-x)} \right]=\frac{\pi }{2}\] \[\Rightarrow \frac{1+x+1-x}{1-(1+x)(1-x)}=\tan \frac{\pi }{2}\] \[\Rightarrow \frac{2}{1-(1+x)(1-x)}=\frac{1}{0}\Rightarrow 1-(1+x)(1-x)=0\] \[\Rightarrow (1+x)(1-x)=1\] \[1-{{x}^{2}}=1\] \[{{x}^{2}}=0\] x = 0
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