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Question:

If $\tan ^{-1} x+\tan ^{-1} \frac{1}{2}=\frac{\pi}{4}$, then $x=$ _________________.

Solution:

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

$\Rightarrow \tan ^{-1}\left(\frac{x+\frac{1}{2}}{1-x \times \frac{1}{2}}\right)=\frac{\pi}{4}$                $\left[\tan ^{-1} a+\tan ^{-1} b=\tan ^{-1}\left(\frac{a+b}{1-a b}\right)\right]$

$\Rightarrow \frac{2 x+1}{2-x}=\tan \frac{\pi}{4}=1$

$\Rightarrow 2 x+1=2-x$

$\Rightarrow 3 x=1$

$\Rightarrow x=\frac{1}{3}$

Thus, the value of $x$ is $\frac{1}{3}$.

If $\tan ^{-1} x+\tan ^{-1} \frac{1}{2}=\frac{\pi}{4}$, then $x=\frac{1}{3}$

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