How to find $ab+cd$ given that $a^2+b^2=c^2+d^2=1$ and $ac+bd=0$?

$$ab+cd=ab(c^2+d^2)+cd(a^2+b^2)=(ad+bc)(ac+bd)=0$$

I assume $a,b \in \mathbb{R}$. Since $a^2+b^2 = 1$, we have $-1 \leq a \leq 1$ and likewise $-1 \leq b \leq 1$. Let us take $a = \cos(\alpha)$ and $b = \sin(\alpha)$ without loss of generality. Similarly, $c = \cos(\beta)$ and $d = \sin(\beta)$.

We have $ac + bd = \sin(\alpha) \sin(\beta) + \cos(\alpha) \cos(\beta) = \cos(\alpha - \beta) = 0$.

You have

$ab+cd = \cos(\alpha) \sin(\alpha) + \cos(\beta) \sin(\beta)\\ = \frac{1}{2} ( \sin(2 \alpha) + \sin(2 \beta) ) \\ = \sin(\alpha+\beta) \cos(\alpha - \beta) \\ = 0 $

The answer should be 0.

You can interptet $(a, b)$ and $(c, d)$ as two orthogonal vectors that lie on the unit circle. Converting this into polar coordinates, this means there are angles $\phi$, $\theta$ such that $(a, b) = (\cos(\phi), \sin(\phi))$, $(c, d) = (\cos(\theta), \sin(\theta))$ and $|\phi - \theta| = \frac{\pi}{2}$. Now observe that $$ab + cd = \cos(\phi)\sin(\phi) + \cos(\theta) \sin(\theta) = \sin(\phi + \theta)\cos(\phi - \theta) = 0$$ because $\cos\left(\pm \frac{\pi}{2}\right) = 0$.