Solution

Let \[ \tan^{-1}\left(\frac{1}{4}\right) = x \] so that \[ \tan x = \frac{1}{4} \]

Then, \[ \sec^2 x = 1 + \tan^2 x = 1 + \frac{1}{16} = \frac{17}{16} \]

Hence, \[ \cos x = \frac{4}{\sqrt{17}}, \quad \sin x = \frac{1}{\sqrt{17}} \]

Now, let \[ \tan^{-1}\left(\frac{2}{9}\right) = y \Rightarrow \tan y = \frac{2}{9} \]

Then, \[ \sec^2 y = 1 + \frac{4}{81} = \frac{85}{81} \] \[ \cos y = \frac{9}{\sqrt{85}}, \quad \sin y = \frac{2}{\sqrt{85}} \]

Using the identity: \[ \sin(x+y) = \sin x \cos y + \cos x \sin y \]

\[ = \frac{1}{\sqrt{17}}\cdot\frac{9}{\sqrt{85}} + \frac{4}{\sqrt{17}}\cdot\frac{2}{\sqrt{85}} \]

\[ = \frac{17}{\sqrt{17}\sqrt{85}} = \frac{1}{\sqrt{5}} \]

\[ \boxed{ \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) } \]