A) \[\frac{3x-{{x}^{3}}}{1+3{{x}^{2}}}\]
B) \[\frac{3x+{{x}^{3}}}{1+3{{x}^{2}}}\]
C) \[\frac{3x-{{x}^{3}}}{1-3{{x}^{2}}}\]
D) \[\frac{3x+{{x}^{3}}}{1-3{{x}^{2}}}\]
Correct Answer: C
Solution :
\[{{\tan }^{-1}}y={{\tan }^{-1}}x+2{{\tan }^{-1}}x\] \[=3{{\tan }^{-1}}x\] \[{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)\] \[y=\frac{3x-{{x}^{3}}}{1-3{{x}^{2}}}\]You need to login to perform this action.
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