A) \[{{\tan }^{-1}}x\]
B) \[\frac{1}{2}{{\tan }^{-1}}x\]
C) \[2{{\tan }^{-1}}x\]
D) None of these
Correct Answer: B
Solution :
\[{{\tan }^{-1}}\left( \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right)={{\tan }^{-1}}\left[ \frac{\sqrt{1+{{\tan }^{2}}\theta }-1}{\tan \theta } \right]\](Putting \[x=\tan \theta )\] \[={{\tan }^{-1}}\left[ \frac{\sec \theta -1}{\tan \theta } \right]={{\tan }^{-1}}\left[ \frac{1-\cos \theta }{\sin \theta } \right]\] \[={{\tan }^{-1}}\left[ \frac{2\,{{\sin }^{2}}\frac{\theta }{2}}{2\,\sin \frac{\theta }{2}\cos \frac{\theta }{2}} \right]\]\[={{\tan }^{-1}}\tan \frac{\theta }{2}=\frac{\theta }{2}=\frac{1}{2}{{\tan }^{-1}}x\].
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