Prove that: $\;$ $\cos^{-1} \left(\dfrac{4}{5}\right) + \sin^{-1} \left(\dfrac{5}{13}\right) = \tan^{-1} \left(\dfrac{56}{33}\right)$
$\cos^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{\sqrt{1 - x^2}}{x}\right)$ $\hspace{1cm}$ if $\;$ $0 < x < 1$
$\sin^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{x}{\sqrt{1 - x^2}}\right)$ $\hspace{1cm}$ if $\;$ $0 < x < 1$
$\begin{aligned}
\therefore \; \cos^{-1} \left(\dfrac{4}{5}\right) + \sin^{-1} \left(\dfrac{5}{13}\right) & = \tan^{-1} \left[\dfrac{\sqrt{1 - \left(\dfrac{4}{5}\right)^2}}{\dfrac{4}{5}}\right] + \tan^{-1} \left[\dfrac{\dfrac{5}{13}}{\sqrt{1 - \left(\dfrac{5}{13}\right)^2}}\right] \\\\
& \left(\text{Note: } 0 < \dfrac{4}{5} < 1, \;\;\; 0 < \dfrac{5}{13} < 1 \right) \\\\
& = \tan^{-1} \left(\dfrac{3}{4}\right) + \tan^{-1} \left(\dfrac{5}{12}\right) \\\\
& \left[\text{Note: } \right. \\
& \left. \tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \tan^{-1} \left(\dfrac{x + y}{1 - x \cdot y}\right), \;\; x \cdot y < 1\right] \\\\
& = \tan^{-1} \left(\dfrac{\dfrac{3}{4} + \dfrac{5}{12}}{1 - \dfrac{3}{4} \times \dfrac{5}{12}}\right) \\\\
& = \tan^{-1} \left(\dfrac{\dfrac{14}{12}}{\dfrac{33}{48}}\right) \\\\
& = \tan^{-1} \left(\dfrac{56}{33}\right)
\end{aligned}$
Hence proved.