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