\[3{{\tan }^{-1}}a\]is equal to [MP PET 1993]

1.	\[3{{\tan }^{-1}}a\]is equal to [MP PET 1993]
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.	\[{{\tan }^{-1}}\frac{3a-{{a}^{3}}}{1-3{{a}^{2}}}\]
Answer» E.

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\[3{{\tan }^{-1}}a\]is equal to [MP PET 1993]

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