Question

The expression ${\tan }^{-1}\left(\frac{6x-8x^3}{1-12x^2}\right)-{\tan }^{-1}\left(\frac{4x}{1-4x^2}\right)$ in the interval $|2x|<\frac{1}{\sqrt{3}}$ simlifies to:

• A. ${\tan }^{-1}2x$
• B. $3x$
• C. $x$
• D. $0$

Answer 1

The correct option is A ${\tan }^{-1}2x$

${\tan }^{-1}\left(\frac{6x-8x^3}{1-12x^2}\right)-{\tan }^{-1}\left(\frac{4x}{1-4x^2}\right)={\tan }^{-1}\left[\frac{3\times 2x-(2x{)}^3}{1-3\times (2x{)}^2}\right]-{\tan }^{-1}\left[\frac{2\times 2x}{1-(2x{)}^2}\right]$
Considering
${\tan }^{-1}2x=\theta \Rightarrow 2x=\tan \theta$
As $|2x|<\frac{1}{\sqrt{3}}\Rightarrow \theta \in (-\frac{\pi }{6},\frac{\pi }{6})$

Now, the expression simplifies to
$={\tan }^{-1}\left[\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right]-{\tan }^{-1}\left[\frac{2\tan \theta }{1-{\tan }^2\theta }\right]={\tan }^{-1}\left(\tan 3\theta \right)-{\tan }^{-1}\left(\tan 2\theta \right)$

In the given interval
$3\theta \in (-\frac{\pi }{2},\frac{\pi }{2}),2\theta \in (-\frac{\pi }{3},\frac{\pi }{3})$
So,
${\tan }^{-1}\left(\tan 3\theta \right)-{\tan }^{-1}\left(\tan 2\theta \right)=\theta ={\tan }^{-1}2x$