Question

If ${\tan }^{-1}\frac{x-3}{x-4}+{\tan }^{-1}\frac{x+3}{x+4}=\frac{\pi }{4}$, then the value of $2x^2$ is

Answer 1

Given : ${\tan }^{-1}\frac{x-3}{x-4}+{\tan }^{-1}\frac{x+3}{x+4}=\frac{\pi }{4}$
Now,
${\tan }^{-1}\frac{x-3}{x-4}={\tan }^{-1}1-{\tan }^{-1}\frac{x+3}{x+4}\Rightarrow {\tan }^{-1}\frac{x-3}{x-4}={\tan }^{-1}\left(\frac{1-\frac{x+3}{x+4}}{1+\frac{x+3}{x+4}}\right)\Rightarrow {\tan }^{-1}\frac{x-3}{x-4}={\tan }^{-1}\frac{1}{2x+7}\Rightarrow \frac{x-3}{x-4}=\frac{1}{2x+7}\Rightarrow 2x^2-6x+7x-21=x-4\therefore 2x^2=17$