Prove that:\(\tan^{-1}\left(\frac{3\mathrm x-\mathrm x^3}{1-3\mathrm x

1.	Prove that:\(\tan^{-1}\left(\frac{3\mathrm x-\mathrm x^3}{1-3\mathrm x^2}\right)=3\tan^{-1}\mathrm x,\|\mathrm x\|<\frac{1}{\sqrt{3}}\)
Answer» To Prove: \(\tan^{-1}\left(\frac{3\mathrm x-\mathrm x^3}{1-3\mathrm x^2}\right)=3\tan^{-1}\mathrm x\) Formula Used: \(\tan3A=\frac{3\tan A-tan^2A}{1-3\tan^2A}\) Proof: LHS \(=\tan^{-1}\left(\frac{3\mathrm x-\mathrm x^2}{1-3\mathrm x^2}\right)\)… (1) Let x = tan A … (2) Substituting (2) in (1), LHS \(=\tan^{-1}\left(\frac{3\tan A-tan^2a}{1-3\tan^2A}\right)\) = tan^-1(tan 3A) = 3A From (2), A = tan ^-1x, 3A = 3 tan ^-1 x = RHS Therefore, LHS = RHS Hence proved.

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

Answer»

To Prove: \(\tan^{-1}\left(\frac{3\mathrm x-\mathrm x^3}{1-3\mathrm x^2}\right)=3\tan^{-1}\mathrm x\)

Formula Used: \(\tan3A=\frac{3\tan A-tan^2A}{1-3\tan^2A}\)

Proof:

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

Let x = tan A … (2)

Substituting (2) in (1),

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

= tan^-1(tan 3A)

= 3A

From (2), A = tan ^-1x,

3A = 3 tan ^-1 x

= RHS

Therefore, LHS = RHS

Hence proved.