Question

If ${\tan }^{-1}\left(\frac{x-1}{x-2}\right)+{\tan }^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi }{4}$, find the value of $x$

Answer 1

We have,

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

We know that

${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}$

Therefore,

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

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

$\frac{x^2+2x-x-2+x^2-2x+x-2}{x^2-4-x^2+1}=1$

$\frac{2x^2-4}{-3}=1$

$2x^2-4=-3$

$2x^2=1$

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

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

Hence, this is the value.