Check the given equality: $\;$ $\sin^{-1} \left(\dfrac{4}{5}\right) - \cos^{-1} \left(\dfrac{2}{\sqrt{5}}\right) = \tan^{-1} \left(\dfrac{1}{2}\right)$
Formulas
$\sin^{-1} x = \tan^{-1} \left(\dfrac{x}{\sqrt{1 - x^2}}\right), \;\;\; -1 < x < 1$
$\cos^{-1} x = \tan^{-1} \left(\dfrac{\sqrt{1 - x^2}}{x}\right), \;\;\; -1 < x < 0, \; 0 < x < 1$
$\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left(\dfrac{x - y}{1 + x \cdot y}\right), \;\;\; x \cdot y < 1$
$\begin{aligned}
LHS & = \sin^{-1} \left(\dfrac{4}{5}\right) - \cos^{-1} \left(\dfrac{2}{\sqrt{5}}\right) \\\\
& = \tan^{-1} \left(\dfrac{\dfrac{4}{5}}{\sqrt{1 - \dfrac{16}{25}}}\right) - \tan^{-1} \left(\dfrac{\sqrt{1 - \dfrac{4}{5}}}{\dfrac{2}{\sqrt{5}}}\right) \\\\
& = \tan^{-1} \left(\dfrac{4}{3}\right) - \tan^{-1} \left(\dfrac{1}{2}\right) \\\\
& = \tan^{-1} \left(\dfrac{\dfrac{4}{3} - \dfrac{1}{2}}{1 + \dfrac{4}{3} \times \dfrac{1}{2}}\right) \\\\
& = \tan^{-1} \left(\dfrac{5}{10}\right) \\\\
& = \tan^{-1} \left(\dfrac{1}{2}\right) = RHS
\end{aligned}$