Question
Solve the following equation for x:
$\tan^{-1}(\text{x}-1)+\tan^{-1}\text{x}+\tan^{-1}(\text{x}+1)=\tan^{-1}3\text{x}$
$\tan^{-1}(\text{x}-1)+\tan^{-1}\text{x}+\tan^{-1}(\text{x}+1)=\tan^{-1}3\text{x}$
✓
Answer
We know
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$ and
$\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)$
$\therefore\ \tan^{-1}(\text{x}+1)+\tan^{-1}(\text{x}-1)+\tan^{-1}\text{x}=\tan^{-1}3\text{x}$
$\Rightarrow\tan^{-1}\Big\{\frac{\text{x}+1+\text{x}-1}{1-(\text{x}+1)\times(\text{x} +1)}\Big\}=\tan^{-1}3\text{x}-\tan^{-1}\text{x}$
$\Rightarrow\tan^{-1}\Big(\frac{2\text{x}}{2-\text{x}^2}\Big)=\tan^{-1}\Big(\frac{3\text{x}-\text{x}}{1+3\text{x}^2}\Big)$
$\Rightarrow\frac{2\text{x}}{2-\text{x}^2}=\frac{2\text{x}}{1+3\text{x}^2}$
$\Rightarrow2-\text{x}^2=1+3\text{x}^2$
$\Rightarrow4\text{x}^2-1=0$
$\Rightarrow\text{x}^2=\frac{1}{4}$
$\Rightarrow\text{x}=\pm\frac{1}{2}$
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$ and
$\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)$
$\therefore\ \tan^{-1}(\text{x}+1)+\tan^{-1}(\text{x}-1)+\tan^{-1}\text{x}=\tan^{-1}3\text{x}$
$\Rightarrow\tan^{-1}\Big\{\frac{\text{x}+1+\text{x}-1}{1-(\text{x}+1)\times(\text{x} +1)}\Big\}=\tan^{-1}3\text{x}-\tan^{-1}\text{x}$
$\Rightarrow\tan^{-1}\Big(\frac{2\text{x}}{2-\text{x}^2}\Big)=\tan^{-1}\Big(\frac{3\text{x}-\text{x}}{1+3\text{x}^2}\Big)$
$\Rightarrow\frac{2\text{x}}{2-\text{x}^2}=\frac{2\text{x}}{1+3\text{x}^2}$
$\Rightarrow2-\text{x}^2=1+3\text{x}^2$
$\Rightarrow4\text{x}^2-1=0$
$\Rightarrow\text{x}^2=\frac{1}{4}$
$\Rightarrow\text{x}=\pm\frac{1}{2}$
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