A) \[1\]
B) \[0\]
C) \[-1\]
D) \[1/\sqrt{2}\]
Correct Answer: A
Solution :
We have, \[{{\sin }^{-1}}\,2x+{{\cos }^{-1}}2x+2{{\tan }^{-1}}x=x\] \[\Rightarrow \] \[\frac{\pi }{2}+2{{\tan }^{-1}}x=\pi \,\left( \because \,\,\,{{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2} \right)\] \[\Rightarrow \] \[2{{\tan }^{-1}}x=\frac{\pi }{1}-\frac{\pi }{2}=\frac{\pi }{2}\] \[\Rightarrow \] \[{{\tan }^{-1}}x=\frac{\pi }{4}\] \[\Rightarrow \] \[x=\tan \frac{\pi }{4}\] \[\Rightarrow \] \[x=1\]
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