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 $\pm \frac{1}{\sqrt{2}}$

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

$⟹{\tan }^{-1}\left(\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{x-1}{x-2}\cdot \frac{x+1}{x+2}}\right)=\frac{\pi }{4}$ ...... [Using formula of ${\tan }^{-1}a+{\tan }^{-1}b$]

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

$⟹\frac{x-1}{x-2}+\frac{x+1}{x+2}=1-\frac{x-1}{x-2}\cdot \frac{x+1}{x+2}$ ....... $[\because \tan \frac{\pi }{4}=1]$

$⟹\frac{x^2+x-2+x^2-x-2}{x^2-4}=1-\frac{x^2-1}{x^2-4}$

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