A) \[0\]
B) \[1\]
C) \[{{\tan }^{-1}}(x)\]
D) \[{{\tan }^{-1}}(2x)\]
Correct Answer: C
Solution :
Let \[x=\tan \theta \Rightarrow \theta ={{\tan }^{-1}}x\] \[\therefore \] \[{{\tan }^{-1}}\left( \frac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)-{{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)\] \[={{\tan }^{-1}}\left( \frac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta } \right)-{{\tan }^{-1}}\left( \frac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right)\] \[={{\tan }^{-1}}(\tan 3\theta )-{{\tan }^{-1}}(\tan 2\theta )\] \[=3\theta -2\theta =\theta ={{\tan }^{-1}}x\]
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