Question

The possible values of $x$, which satisfy the trigonometric equation ${\tan }^{-1}\left(\frac{x-1}{x-2}\right)+{\tan }^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi }{4}$ are

• A. $\pm \frac{1}{\sqrt{2}}$
• B. $\pm \sqrt{2}$
• C. $\pm \frac{1}{2}$
• D. $\pm 2$

Answer 1

The correct option is A $\pm \frac{1}{\sqrt{2}}$

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

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

$\Rightarrow \frac{x-1}{x-2}+\frac{x+1}{x+2}=1-\frac{x-1}{x-2}\times \frac{x+1}{x+2}\Rightarrow x^2+x-2+x^2-x-2=x^2-4-x^2+1\Rightarrow 2x^2=1\Rightarrow x=\pm \frac{1}{\sqrt{2}}$