Question

Find the value of $x$
If,$ta{n}^{-1}(x-1)+ta{n}^{-1}x+ta{n}^{-1}(x+1)=ta{n}^{-1}3x$

Answer 1

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

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

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

$\Rightarrow \frac{2x-1}{1-x^2+x}=\frac{2x-1}{1+3x^2+3x}$

$\Rightarrow (2x-1)(1+3x^2+3x)=(1-x^2+x)(2x-1)$

$\Rightarrow (2x-1)[(1+3x^2+3x)-(1-x^2+x)]=0$

$\Rightarrow (2x-1)[1+3x^2+3x-1+x^2-x]=0$

$\Rightarrow (2x-1)[4x^2+2x]=0$

$\Rightarrow (2x-1)2x(2x+1)=0$

$\Rightarrow (2x-1)=0,2x=0,(2x+1)=0$

$\Rightarrow 2x=1,x=0,2x=-1$

$\Rightarrow x=\frac{1}{2},x=0,x=\frac{-1}{2}$

$\therefore x=0,\pm \frac{1}{2}$