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

Answer: (A)

We have,
$\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$
$\Rightarrow \tan ^{-1}\left[\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{x-1}{x-2} \cdot \frac{x+1}{x+2}}\right]=\frac{\pi}{4}$
$\Rightarrow \frac{(x-1)(x+2)+(x-2)(x+1)}{(x-2)(x+2)-(x-1)(x+1)}=\tan \frac{\pi}{4} $
$\Rightarrow \frac{x^{2}+x-2+x^{2}-x-2}{x^{2}-4-x^{2}+1}=1$
$ \Rightarrow \frac{2 x^{2}-4}{-3} =1 $
$\Rightarrow 2 x^{2}-4 =-3$
$ \Rightarrow 2 x^{2} =1$
$\Rightarrow x^{2} =\frac{1}{2} $
$ \Rightarrow x =\pm \frac{1}{\sqrt{2}} $