Question

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

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

Answer 1

Answer: (A)

We have,
${\tan }^{−1}\left(\frac{x−1}{x−2}\right)+{\tan }^{−1}\left(\frac{x+1}{x+2}\right)=\frac{π}{4}$
$⇒{\tan }^{−1}\left[\frac{\frac{x−1}{x−2}+\frac{x+1}{x+2}}{1−\frac{x−1}{x−2}â‹\dots \frac{x+1}{x+2}}\right]=\frac{Ï€}{4}$
$⇒\frac{(x−1)(x+2)+(x−2)(x+1)}{(x−2)(x+2)−(x−1)(x+1)}=\tan \frac{π}{4}$
$⇒\frac{x^2+x−2+x^2−x−2}{x^2−4−x^2+1}=1$
$⇒\frac{2x^2−4}{−3}=1$
$⇒2x^2−4=−3$
$⇒2x^2=1$
$⇒x^2=\frac{1}{2}$
$⇒x=Â\pm \frac{1}{\sqrt{2}}$