First, if $\alpha+\beta=\frac{\pi}{2}$, then $$\sin\alpha=\cos\beta \implies \sin^2 \alpha = \cos^2 \beta$$ and $$\cos\alpha=\sin\beta \implies \cos^2\alpha = \sin^2\beta$$ Hence $$\frac{\sin^4 \alpha}{\cos^2 \beta} + \frac{\cos^4\alpha}{\sin^2\beta} = \sin^2\alpha + \cos^2\alpha= 1$$ Now, let's prove that if $\frac{\sin^4\alpha}{\cos^2\beta} + \cos^2\beta = 1$, then $\alpha + \beta=\frac{\pi}{2}$. $\displaystyle\frac{\sin^4\alpha}{\cos^2\beta} + \cos^2\beta \ge 2\sin^2\alpha\quad\text{and}\quad\displaystyle\frac{\cos^4\alpha}{\sin^2\beta} + \sin^2\beta \ge 2\cos^2\alpha$ Hence $$\frac{\sin^4 \alpha}{\cos^2 \beta} + \frac{\cos^4\alpha}{\sin^2\beta} \ge 1$$ Therefore $$\sin^2\alpha = \cos^2\beta\quad\text{and}\quad\cos^2\alpha = \sin^2\beta$$ It follows that $$ \begin{array}{rl} \cos^2\alpha\cos^2\beta-\sin^2\alpha\sin^2\beta &=0\\ (\cos\alpha\cos\beta+\sin\alpha\sin\beta) (\cos\alpha\cos\beta-\sin\alpha\sin\beta) &=0\\ \cos(\alpha+\beta) \cos(\alpha-\beta) &=0\\ \end{array} $$ Because $\alpha, \beta \in (0, \frac{\pi}{2})$, $-\frac{\pi}{2} < \alpha - \beta < \frac{\pi}{2}$ and $0 < \alpha + \beta < \pi$. So we conclude $\alpha + \beta =\frac{\pi}{2}$.