Solution:

$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}$

$\quad=\tan ^{-1}\left[\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} \times \frac{2}{9}}\right]$

$$\begin{aligned} &=\tan ^{-1}\left(\frac{1}{2}\right) \ &=\cot ^{-1} 2 \ \therefore & \tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\cot ^{-1} 2 \end{aligned}$$