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
$tan3A=\frac{3tanA-ta{n}^{3}A}{1-3ta{n}^{2}A}$
$\therefore 3A=ta{n}^{-1}\frac{3tanA-ta{n}^{3}A}{1-3ta{n}^{2}A}$
Put $tanA=a$ i.e., $A=ta{n}^{-1}a$
$\therefore 3ta{n}^{-1}a=ta{n}^{-1}\frac{3a-a^3}{1-3a^2}$