Mark the tick against the correct answer in the following:

To Find: The value of $\tan ^{-1} x=\frac{\pi}{4}-\tan ^{-1} \frac{1}{3}$

Now, $\tan ^{-1} x=\tan ^{-1} 1-\tan ^{-1} \frac{1}{3}\left(\because \tan \frac{\pi}{4}=1\right)$

Since we know that $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$

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

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

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