\(\tan ^{-1}(1 / 4)+\tan ^{-1}(2 / 9)=\) (a) \((1 / 2) \cos ^{-1}(3 / 5)\) (b) \((1 / 2) \sin ^{-1}(4 / 5)\) (c) \((1 / 2) \tan ^{-1}(3 / 5)\) (d) \(\tan ^{-1}(8 / 9)\)

The tangent addition formula is given by: \[\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}\] We will use this formula to find the tangent of the sum of the given angles, i.e., \(\tan\left(\tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right)\right)\).

Applying the tangent addition formula, we have: \[\tan\left(\tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right)\right) = \frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4}\cdot\frac{2}{9}}\]

By simplifying the expression, we get: \[\tan\left(\tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right)\right) = \frac{\frac{9 + 8}{36}}{1 - \frac{1}{18}} = \frac{17}{36}\div\frac{17}{18}\] \[\tan\left(\tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right)\right) = \frac{17}{36} \cdot \frac{18}{17} = \frac{18}{36}\] \[\tan\left(\tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right)\right) = \frac{1}{2}\]

Now, we need to compare our result with the given options to find the correct answer: (a) \(\frac{1}{2}\cos^{-1}\left(\frac{3}{5}\right)\) is incorrect because the result is not in terms of cosine. (b) \(\frac{1}{2}\sin^{-1}\left(\frac{4}{5}\right)\) is incorrect because the result is not in terms of sine. (c) \(\frac{1}{2}\tan^{-1}\left(\frac{3}{5}\right)\) is incorrect because the result is not \(\frac{1}{2}\tan^{-1}\left(\frac{3}{5}\right)\), but just \(\frac{1}{2}\). (d) \(\tan^{-1}\left(\frac{8}{9}\right)\) is incorrect because the result is not in terms of arctangent of a single value. Since none of the given options correctly represent our result, we can conclude that this exercise does not have a valid answer among the given options.