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Question:

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

Solution:

$\tan ^{-1} x=\frac{\pi}{4}-\tan ^{-1} \frac{1}{3}$

$\Rightarrow \tan ^{-1} x=\tan ^{-1} 1-\tan ^{-1} \frac{1}{3}$

$\Rightarrow \tan ^{-1} x=\tan ^{-1}\left(\frac{1-\frac{1}{3}}{1+1 \times \frac{1}{3}}\right)$                $\left[\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right), x y>-1\right]$

$\Rightarrow \tan ^{-1} x=\tan ^{-1}\left(\frac{\frac{2}{3}}{\frac{4}{3}}\right)$

$\Rightarrow \tan ^{-1} x=\tan ^{-1} \frac{1}{2}$

$\Rightarrow x=\frac{1}{2}$

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

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