Question

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

Answer 1

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

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

Taking tangent on both sides $\frac{\frac{x-2}{x-4}+\frac{x+2}{x+4}}{1-\frac{x-2}{x-4}\times \frac{x+2}{x+4}}=tan\left(\frac{\pi }{4}\right)$

$\Rightarrow \frac{(x-2)(x+4)+(x+2)(x-4)}{x^2-16-(x^2-4)}=1$

$\Rightarrow \frac{x^2+2x-8+(x^2-2x-8)}{-12}=1$

$\Rightarrow 2x^2-16=-12$

$\Rightarrow 2x^2=4\Rightarrow x^2=2\Rightarrow x=\pm \sqrt{2}$