Question

$3{\tan }^{-1}\left(a\right)=?$

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

Answer 1

The correct option is D

${\tan }^{-1}\left(\frac{3a-a^3}{1-3a^2}\right)$

Explanation for the correct option:

Finding the value.

Given, $3{\tan }^{-1}\left(a\right)$

We know that,

$\tan \left(3a\right)=\frac{3\tan \left(a\right)-{\tan }^3\left(a\right)}{1-3{\tan }^2\left(a\right)}$

$\Rightarrow 3a={\tan }^{-1}\left(\frac{3\tan \left(a\right)-{\tan }^3\left(a\right)}{1-3{\tan }^2\left(a\right)}\right)$

Let, $\tan \left(a\right)=a\Rightarrow a={\tan }^{-1}\left(a\right)$

$\Rightarrow 3{\tan }^{-1}\left(a\right)={\tan }^{-1}\left(\frac{3a-a^3}{1-3a^2}\right)$

Therefore, correct answer is option $\left(D\right)$