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Prove that:

Question:

Prove that:

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

 

Solution:

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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