If the inverse trigonometric functions take principal values, then
$$\cos^{-1}\left(\frac{3}{10}\cos\left(\tan^{-1}\left(\frac{4}{3}\right)\right) + \frac{2}{5}\sin\left(\tan^{-1}\left(\frac{4}{3}\right)\right)\right)$$ is equal to

We need to evaluate $$\cos^{-1}\left(\frac{3}{10}\cos\left(\tan^{-1}\frac{4}{3}\right) + \frac{2}{5}\sin\left(\tan^{-1}\frac{4}{3}\right)\right)$$. First, let $$\theta = \tan^{-1}(4/3)$$ so that $$\tan\theta = 4/3$$. In a right triangle with opposite = 4, adjacent = 3, and hypotenuse = 5, it follows that $$\cos\theta = \frac{3}{5}$$ and $$\sin\theta = \frac{4}{5}$$.

Next, substituting these values into the expression gives $$\frac{3}{10}\cdot\frac{3}{5} + \frac{2}{5}\cdot\frac{4}{5} = \frac{9}{50} + \frac{8}{25} = \frac{9}{50} + \frac{16}{50} = \frac{25}{50} = \frac{1}{2}$$.

Now, since $$\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$$, the answer is Option C: $$\dfrac{\pi}{3}$$.