Question

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

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,

${\tan }^{-1}\left(m\right)+{\tan }^{-1}\left(n\right)={\tan }^{-1}\left(\frac{m+n}{1-mn}\right)$

Therefore,

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

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

or, $\frac{(x-1)(x+2)+(x+1)(x-2)}{x^2-4-(x^2-1)}=1$

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

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

or, $2x^2-4=-3$

or, $2x^2=1$

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

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