If the inverse trigonometric functions take principal values, then...

Let $\tan^{- 1}\left(\dfrac{4}{3}\right) = \theta ⇒ tan\theta = \dfrac{4}{3}$

$∴ cos\theta = \dfrac{3}{5} \text{ and } sin\theta = \dfrac{4}{5}$

So, $\cos^{- 1}\left\{\dfrac{3}{10}\cos \left(\tan^{- 1}\dfrac{4}{3}\right) + \dfrac{2}{5}\sin \left(\tan^{- 1}\dfrac{4}{3}\right)\right\}$

$= \cos^{- 1}\left\{\dfrac{3}{10}cos\theta + \dfrac{2}{5}sin\theta\right\}$

$= \cos^{- 1}\left\{\dfrac{3}{10} \times \dfrac{3}{5} + \dfrac{2}{5} \times \dfrac{4}{5}\right\}$

$= \cos^{- 1}\left\{\dfrac{9}{50} + \dfrac{8}{25}\right\}$

$= \cos^{- 1}\dfrac{1}{2} = \dfrac{\pi}{3}$