Question

Prove that
${\tan }^{-1}\left(\frac{6x-8x^3}{1-12x^2}\right)-{\tan }^{-1}\left(\frac{4x}{1-4x^2}\right)={\tan }^{-1}2x;|2x|<\frac{1}{\sqrt{3}}$.

Answer 1

L.H.S = ${\tan }^{-1}\left(\frac{6x-8x^3}{1-12x^2}\right)-{\tan }^{-1}\left(\frac{4x}{1-4x^2}\right)$

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

Considering ${\tan }^{-1}2x=\theta \text{then}2x=\tan \theta$

Now,

= ${\tan }^{-1}\left[\frac{3\tan \theta -(\tan \theta {)}^3}{1-3(\tan \theta {)}^2}\right]-{\tan }^{-1}\left[\frac{2\tan \theta }{1-{\tan }^2\theta }\right]$

= ${\tan }^{-1}\left[\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right]-{\tan }^{-1}\left[\frac{2\tan \theta }{1-{\tan }^2\theta }\right]$

= ${\tan }^{-1}\left(\tan 3\theta \right)-{\tan }^{-1}\left(\tan 2\theta \right)$

= $3\theta -2\theta =\theta ={\tan }^{-1}2x$ = R.H.S.

Hence proved.