Check the given equality: $\;$ $\sin^{-1} \left(\dfrac{7}{25}\right) + \dfrac{1}{2} \cos^{-1} \left(\dfrac{7}{25}\right) = \cos^{-1} \left(\dfrac{3}{5}\right)$
To Check That (TCT) $\;$ $\sin^{-1} \left(\dfrac{7}{25}\right) + \dfrac{1}{2} \cos^{-1} \left(\dfrac{7}{25}\right) = \cos^{-1} \left(\dfrac{3}{5}\right)$
i.e. TCT $\;$ $\left[\sin^{-1} \left(\dfrac{7}{25}\right) + \cos^{-1} \left(\dfrac{7}{25}\right)\right] - \dfrac{1}{2} \cos^{-1} \left(\dfrac{7}{25}\right) = \cos^{-1} \left(\dfrac{3}{5}\right)$
i.e. TCT $\;$ $\dfrac{\pi}{2} = \cos^{-1} \left(\dfrac{3}{5}\right) + \dfrac{1}{2} \cos^{-1} \left(\dfrac{7}{25}\right)$
$\left\{\because \;\; \sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2}, \;\;\; \forall \; x \in \left[-1, 1\right]\right\}$
i.e. TCT $\;$ $\pi = 2 \cos^{-1} \left(\dfrac{3}{5}\right) + \cos^{-1} \left(\dfrac{7}{25}\right)$
$\begin{aligned}
RHS & = 2 \cos^{-1} \left(\dfrac{3}{5}\right) + \cos^{-1} \left(\dfrac{7}{25}\right) \\\\
& = \cos^{-1} \left(2 \times \dfrac{9}{25} - 1\right) + \cos^{-1} \left(\dfrac{7}{25}\right) \\
& \left\{\because \; 2 \cos^{-1} x = \cos^{-1} \left(2 x^2 - 1\right), \;\; 0 \leq x \leq 1\right\} \\\\
& = \cos^{-1} \left(\dfrac{-7}{25}\right) + \cos^{-1} \left(\dfrac{7}{25}\right) \\\\
& = \pi - \cos^{-1} \left(\dfrac{7}{25}\right) + \cos^{-1} \left(\dfrac{7}{25}\right) \\
& \left\{\because \; \cos^{-1} \left(-x\right) = \pi - \cos^{-1} \left(x\right) \;\; \forall \;\; x \in \left[-1, 1\right]\right\} \\\\
& = \pi = LHS
\end{aligned}$