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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
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$\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$ are
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Verified Answer
The correct answer is:
$\pm \frac{1}{\sqrt{2}}$
We have, $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$
$\Rightarrow \quad \tan ^{-1}\left[\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{x-1}{x-2} \cdot \frac{x+1}{x+2}}\right]=\frac{\pi}{4}$
$\Rightarrow \frac{(x-1)(x+2)+(x-2)(x+1)}{(x-2)(x+2)-(x-1)(x+1)}=\tan \frac{\pi}{4}$
$\Rightarrow \quad \frac{x^{2}+x-2+x^{2}-x-2}{x^{2}-4-x^{2}+1}=1$
$\Rightarrow$ $\frac{2 x^{2}-4}{-3}=1$
$\Rightarrow$ $2 x^{2}-4=3$
$\Rightarrow$ $2 x^{2}=1$
$x^{2}=\frac{1}{2}$
$x=\pm \frac{1}{\sqrt{2}}$
$\Rightarrow \quad \tan ^{-1}\left[\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{x-1}{x-2} \cdot \frac{x+1}{x+2}}\right]=\frac{\pi}{4}$
$\Rightarrow \frac{(x-1)(x+2)+(x-2)(x+1)}{(x-2)(x+2)-(x-1)(x+1)}=\tan \frac{\pi}{4}$
$\Rightarrow \quad \frac{x^{2}+x-2+x^{2}-x-2}{x^{2}-4-x^{2}+1}=1$
$\Rightarrow$ $\frac{2 x^{2}-4}{-3}=1$
$\Rightarrow$ $2 x^{2}-4=3$
$\Rightarrow$ $2 x^{2}=1$
$x^{2}=\frac{1}{2}$
$x=\pm \frac{1}{\sqrt{2}}$
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