If $\sin^{-1}(x) + \sin^{-1}(y) = \dfrac{\pi}{2}$, then $x^2 + y^2$ equals:

Let $\sin^{-1}(x) = \alpha$ and $\sin^{-1}(y) = \beta$. So $\alpha + \beta = \dfrac{\pi}{2} \Rightarrow \beta = \dfrac{\pi}{2} - \alpha$. Then $y = \sin\beta = \sin\left(\dfrac{\pi}{2} - \alpha\right) = \cos\alpha$. Since $\sin^2\alpha + \cos^2\alpha = 1$: $x^2 + y^2 = \sin^2\alpha + \cos^2\alpha = 1$.