Question

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

• A. $\tan\frac{3a+a^3}{1+3a^2}$
• B. $\tan^{-1}\frac{3a-a^3}{1+3a^2}$
• C. $\tan^{-1}\frac{3a+a^3}{1-3a^2}$
• D. $\tan^{-1}\frac{3a-a^3}{1-3a^2}$

Answer 1

Answer: (D)

We know that
$ tan \,3A =\frac{ 3\,tan\,A-tan^{3}A}{1-3\,tan^{2}A}$
$\therefore 3\,A = tan^{-1} \frac{3\,tan\,A-tan^{3}A}{1-3tan^{2}A} $
Put $tan\, A = a$ i.e., $A = tan^{-1}a $
$ \therefore 3tan^{-1} a = tan^{-1} \frac{3a-a^{3}}{1-3a^{2}} $