inv-trig

Question: $3{{\tan }^{-1}}a$is equal to



1) ${{\tan }^{-1}}\frac{3a+{{a}^{3}}}{1+3{{a}^{2}}}$
2) ${{\tan }^{-1}}\frac{3a-{{a}^{3}}}{1+3{{a}^{2}}}$
3) ${{\tan }^{-1}}\frac{3a+{{a}^{3}}}{1-3{{a}^{2}}}$
4) ${{\tan }^{-1}}\frac{3a-{{a}^{3}}}{1-3{{a}^{2}}}$
Solution: Explanation: No Explanation
Inverse trigonometric functions

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