Question

 

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

 

Answer 1

 

Given that

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

Taking LHS, we get:

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

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

${\displaystyle ={\tan }^{-1}\left(\left[\frac{(x-2)(x+4)(x+2)(x-4)}{x^2-16-(x^2-4)}\right]\right)}$

${\displaystyle ={\tan }^{-1}\left[\frac{x^2+2x-8+x^2-2x-8}{12}\right]}$

${\displaystyle ={\tan }^{-1}\left[\frac{2x^2-16}{-12}\right]}$

hence

${\displaystyle {\tan }^{-1}\left[\frac{2x^2-16}{-12}\right]=\frac{\pi }{4}}$

${\displaystyle \left[\frac{2x^2-16}{-12}\right]=\tan \left(\frac{\pi }{4}\right)}$

${\displaystyle \Rightarrow \frac{x^2-8}{-6}=1}$

${\displaystyle \Rightarrow x^2-8=-6}$

${\displaystyle \Rightarrow x^2=2}$

${\displaystyle \Rightarrow x=\pm 2}$