A) \[{{\tan }^{-1}}\frac{3a+{{a}^{3}}}{1+3{{a}^{2}}}\]
B) \[{{\tan }^{-1}}\frac{3a-{{a}^{3}}}{1+3{{a}^{2}}}\]
C) \[{{\tan }^{-1}}\frac{3a+{{a}^{3}}}{1-3{{a}^{2}}}\]
D) \[{{\tan }^{-1}}\frac{3a-{{a}^{3}}}{1-3{{a}^{2}}}\]
Correct Answer: D
Solution :
It is obvious.
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