Check the given equality: $\;$ $\tan^{-1} \left(\dfrac{\sqrt{2}}{2}\right) + \sin^{-1} \left(\dfrac{\sqrt{2}}{2}\right) = \tan^{-1} \left(3 + 2 \sqrt{2}\right)$
$\begin{aligned} LHS & = \tan^{-1} \left(\dfrac{\sqrt{2}}{2}\right) + \sin^{-1} \left(\dfrac{\sqrt{2}}{2}\right) \\\\ & = \tan^{-1} \left(\dfrac{1}{\sqrt{2}}\right) + \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right) \\\\ & = \tan^{-1} \left(\dfrac{1}{\sqrt{2}}\right) + \tan^{-1} \left(\dfrac{\dfrac{1}{\sqrt{2}}}{\sqrt{1 - \dfrac{1}{2}}}\right) \\ & \left\{\because \;\; \sin^{-1} x = \tan^{-1} \left(\dfrac{x}{\sqrt{1 - x^2}}\right), \; 0 < x < 1\right\} \\\\ & = \tan^{-1} \left(\dfrac{1}{\sqrt{2}}\right) + \tan^{-1} \left(1\right) \\\\ & = \tan^{-1} \left[\dfrac{\dfrac{1}{\sqrt{2}} + 1}{1 - \dfrac{1}{\sqrt{2}} \times 1}\right] \\ & \left\{\because \;\; \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left(\dfrac{x + y}{1 - xy}\right), \; x \cdot y < 1\right\} \\\\ & = \tan^{-1} \left[\dfrac{\sqrt{2} + 1}{\sqrt{2} - 1}\right] \\\\ & = \tan^{-1} \left[\dfrac{\left(\sqrt{2} + 1\right)^2}{\left(\sqrt{2} - 1 \right) \left(\sqrt{2} + 1\right)}\right] \\\\ & = \tan^{-1} \left[\dfrac{2 + 1 + 2 \sqrt{2}}{2 - 1}\right] \\\\ & = \tan^{-1} \left[3 + 2 \sqrt{2}\right] = RHS \end{aligned}$