Question

Solve for x: ${\tan }^{-1}(x-1)+{\tan }^{-1}x+{\tan }^{-1}(x+1)={\tan }^{-1}3x.$

Answer 1

We have

${\tan }^{-1}(x-1)+{\tan }^{-1}x+{\tan }^{-1}(x+1)={\tan }^{-1}3x$ ...............(1)

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

$[\text{Using}{\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{\left(A\right)+\left(B\right)}{1-\left(A\right)\left(B\right)}\right)]$

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

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

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

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

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

$\Rightarrow {\tan }^{-1}\left[\frac{4x-x^3}{2-3x^2}\right]={\tan }^{-1}3x$

$\Rightarrow \frac{4x-x^3}{2-3x^2}=3x$

$\Rightarrow 4x-x^3=6x-9x^3$

$\Rightarrow 8x^3-2x=0$

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

$\Rightarrow x=0\text{or}4x^2-1=0$

$\Rightarrow x=0\text{or}4x^2=1$

$\Rightarrow x=0\text{or}{x}^2=\frac{1}{4}$

$\Rightarrow x=0\text{or}x=\pm \frac{1}{2}$

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