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.You need to login to perform this action.
You will be redirected in
3 sec