Prove that: $\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)=3 \tan ^{-1} x,|x|

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Prove that: $\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)=3 \tan ^{-1} x,|x|<\frac{1}{\sqrt{3}}$

Prove that: $\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)=3 \tan ^{-1} x,|x|<\frac{1}{\sqrt{3}}$

Solution:

To Prove: $\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)=3 \tan ^{-1} x$
Formula Used: $\tan 3 A=\frac{3 \tan A-\tan ^{3} A}{1-3 \tan ^{2} A}$
Proof:
$\text { LHS }=\tan ^{-1}\left(\frac{3 x-x^{8}}{1-3 x^{2}}\right) \ldots \text { (1) }$
Let $x=\tan A \ldots$ (2)
Substituting (2) in (1),
$\begin{array}{l} \text { LHS }=\tan ^{-1}\left(\frac{3 \tan A-\tan ^{3} A}{1-\tan ^{2} A}\right) \\ =\tan ^{-1}(\tan 3 A) \\ =3 A \end{array}$
From (2), $A=\tan ^{-1} x$
$\begin{array}{l} 3 A=3 \tan ^{-1} x \\ =\text { RHS } \end{array}$
Therefore, LHS = RHS
Hence proved.