Question

If ${\tan }^{-1}\frac{x-3}{x-4}+{\tan }^{-1}\frac{x+3}{x+4}=\frac{\pi }{4}$, then find the value of $x$.

Answer 1

Given ${\tan }^{-1}\frac{x-3}{x-4}+{\tan }^{-1}\frac{x+3}{x+4}=\frac{\pi }{4}$

$\Rightarrow {\tan }^{-1}\left[\begin{array}{c}\frac{\frac{x-3}{x-4}+\frac{x+3}{x+4}}{1-\frac{x-3}{x-4}\times \frac{x+3}{x+4}}\end{array}\right]=\frac{\pi }{4}$

$\Rightarrow \frac{(x-3)(x+4)+(x+3)(x-4)}{(x-4)(x+4)-(x-3)(x+3)}=\tan \frac{\pi }{4}$

$\Rightarrow \frac{x^2+x-12+x^2-x-12}{x^2-16-x^2+9}=1$

$\Rightarrow \frac{2x^2-24}{-7}=1$

$\Rightarrow 2x^2=17$

$\Rightarrow x=\pm \sqrt{\frac{17}{2}}$