Prove that:

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:

$\mathrm{LHS}=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right) \cdots$

Let $x=\tan A \ldots(2)$

Substituting (2) in (1),

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

$=\tan ^{-1}(\tan 3 \mathrm{~A})$

$=3 \mathrm{~A}$

From (2), $A=\tan ^{-1} x$

$3 \mathrm{~A}=3 \tan ^{-1} \mathrm{x}$

$=\mathrm{RHS}$

Therefore, LHS $=$ RHS

Hence proved.