If $\alpha = \cos^{-1} \left(\dfrac{4}{5}\right)$, $\;$ $\beta = \tan^{-1} \left(\dfrac{2}{3}\right)$, $\;$ $\alpha, \; \beta \; \in \; \left(0, \dfrac{\pi}{2}\right)$, $\;$ then show that $\;$ $\left(\alpha - \beta\right) = \tan^{-1} \left(\dfrac{1}{18}\right)$
Given: $\;$ $\alpha = \cos^{-1} \left(\dfrac{4}{5}\right)$, $\;$ $\beta = \tan^{-1} \left(\dfrac{2}{3}\right)$
$\implies$ $\cos \alpha = \dfrac{4}{5}$, $\;$ $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \dfrac{16}{25}} = \dfrac{3}{5}$, $\;$ $\tan \alpha = \dfrac{3}{4}$
and, $\;$ $\tan \beta = \dfrac{2}{3}$
Now, $\;$ $\tan \left(\alpha - \beta\right) = \dfrac{\tan \alpha - \tan \beta}{1 + \tan \alpha \; \tan \beta}$
i.e. $\;$ $\tan \left(\alpha - \beta\right) = \dfrac{\dfrac{3}{4} - \dfrac{2}{3}}{1 + \dfrac{3}{4} \times \dfrac{2}{3}} = \dfrac{1}{18} $
i.e. $\;$ $\left(\alpha - \beta\right) = \tan^{-1} \left(\dfrac{1}{18}\right)$