Solution:

Option(A) is correct.
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 $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$
$\begin{array}{l}
\Rightarrow \tan ^{-1} 1+\tan ^{-1} \frac{1}{3}=\tan ^{-1}\left(\frac{1-\frac{1}{2}}{1+\frac{1}{2}}\right)=\tan ^{-1} \frac{1}{2} \\
\Rightarrow \tan ^{-1} x=\tan ^{-1} \frac{1}{2} \\
\Rightarrow x=\frac{1}{2}
\end{array}$