Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

The given equation can be written as

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

$\Rightarrow 4{\mathrm{x}}^3-\mathrm{x}=0\Rightarrow \mathrm{x}\left(4{\mathrm{x}}^2-1\right)=0$

$\Rightarrow \mathrm{x}=0,\mathrm{x}=\pm \frac{1}{2}$. Thus $\mathrm{x}=\frac{1}{2}$