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