Solution:

Option(B) is correct.
To Find: The value of $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}$
Let, $x=\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}$
Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$\Rightarrow \tan ^{-1} 1+\tan ^{-1} \frac{1}{3}=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{3} \times \frac{3}{2}}\right)=\tan ^{-1} 1=\frac{\pi}{4}$