Question

Find the value of $x$ in given equation ${\tan }^{-1}\left(\frac{x-1}{x-2}\right)+{\tan }^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi }{4}$

Answer 1

Consider the given equation

${\tan }^{-1}\left(\frac{x-1}{x-2}\right)+{\tan }^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi }{4}$

We know that

${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

Therefore,

${\tan }^{-1}\left(\frac{\left(\frac{x-1}{x-2}\right)+\left(\frac{x+1}{x+2}\right)}{1-\left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)}\right)=\frac{\pi }{4}$

${\tan }^{-1}\left(\frac{\frac{(x-1)(x+2)+(x+1)(x-2)}{x^2-4}}{1-\frac{x^2-1}{x^2-4}}\right)=\frac{\pi }{4}$

${\tan }^{-1}\left(\frac{\frac{x^2+2x-x-2+x^2-2x+x-2}{x^2-4}}{\frac{x^2-4-x^2+1}{x^2-4}}\right)=\frac{\pi }{4}$

$ta{n}^{-1}\left(\frac{2x^2-4}{-4+1}\right)=\frac{\pi }{4}$

$\left(\frac{2x^2-4}{-3}\right)=\tan \frac{\pi }{4}$

$2x^2-4=-3$

$2x^2=1$

$x^2=\frac{1}{2}$

$x=\pm \frac{1}{\sqrt{2}}$

Hence, the value of $x$ is $\pm \frac{1}{\sqrt{2}}$.