Prove that: $\;$ $2 \sin^{-1} \left(\dfrac{3}{5}\right) + \cos^{-1} \left(\dfrac{24}{25}\right) = \dfrac{\pi}{2}$
$\sin^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{x}{\sqrt{1 - x^2}}\right)$ $\hspace{1cm}$ if $\;$ $0 < x < 1$
$\cos^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{\sqrt{1 - x^2}}{x}\right)$ $\hspace{1cm}$ if $\;$ $0 < x < 1$
$\begin{aligned}
\therefore \; 2 \sin^{-1} \left(\dfrac{3}{5}\right) + \cos^{-1} \left(\dfrac{24}{25}\right) & = 2 \tan^{-1} \left[\dfrac{\dfrac{3}{5}}{\sqrt{1 - \left(\dfrac{3}{5}\right)^2}}\right] + \tan^{-1} \left[\dfrac{\sqrt{1 - \left(\dfrac{24}{25}\right)^2}}{\dfrac{24}{25}}\right] \\\\
& \left(\text{Note: } 0 < \dfrac{3}{5} < 1, \;\;\; 0 < \dfrac{24}{25} < 1 \right) \\\\
& = 2 \tan^{-1} \left(\dfrac{3}{4}\right) + \tan^{-1} \left(\dfrac{7}{24}\right) \\\\
& \left[\text{Note: } 2 \tan^{-1} \left(x\right) = \tan^{-1} \left(\dfrac{2x}{1 - x^2}\right), \;\; x^2 < 1\right] \\\\
& = \tan^{-1} \left[\dfrac{2 \times \dfrac{3}{4}}{1 - \left(\dfrac{3}{4}\right)^2}\right] + \tan^{-1} \left(\dfrac{7}{24}\right) \\\\
& = \tan^{-1} \left(\dfrac{24}{7}\right) + \tan^{-1} \left(\dfrac{7}{24}\right) \\\\
& \left[\text{Note: } \tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \dfrac{\pi}{2}, \;\; \text{if } x \cdot y = 1\right] \\\\
& = \dfrac{\pi}{2} \;\;\; \left[\because \dfrac{24}{7} \times \dfrac{7}{24} = 1\right]
\end{aligned}$
Hence proved.