If $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$, then the…
If $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$, then the values of $x$ are
- $\pm \frac{3}{\sqrt{2}}$
- $\pm \frac{1}{2}$
- $\pm \frac{1}{\sqrt{2}}$
- $\pm \frac{\sqrt{3}}{2}$
Solution
$\begin{aligned}
& \tan ^{-1}\left(\frac{x-1}{x+2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4} \\
& \therefore \tan ^{-1}\left[\frac{\left(\frac{x-1}{x-2}\right)+\left(\frac{x+1}{x+2}\right)}{1-\left(\frac{x-1}{x-2}\right)+\left(\frac{x+1}{x+2}\right)}\right]=\frac{\pi}{4} \\
& \therefore \frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x-1)(x+1)}=\tan \frac{\pi}{4} \\
& \therefore \frac{\left(x^2+x-2\right)+\left(x^2-x-2\right)}{\left(x^2-4\right)-\left(x^2-1\right)}=1 \\
& \therefore 2 x^2-4=-3 \Rightarrow 2 x^2=1
\end{aligned}$
$\Rightarrow \mathrm{x}= \pm \frac{1}{\sqrt{2}}$
Asked in: MHT CET 2021 (21 Sep Shift 1)
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