Prove that:
$\tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$
To Prove: $\tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$
Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
Proof:
$\mathrm{LHS}=\tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \ldots$ (1)
$=\tan ^{-1}\left(\frac{x+\left(\frac{2 x}{1-x^{2}}\right)}{1-\left(x \times\left(\frac{2 x}{1-x^{2}}\right)\right)}\right)$
$=\tan ^{-1}\left(\frac{x\left(1-x^{2}\right)+2 x}{1-x^{2}-2 x^{2}}\right)$
$=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$
= RHS
Therefore, LHS $=$ RHS
Hence proved.
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