2021/057) Given $a^2+b^2=c^2+d^2=1$ and ac+bd=0.Compute the value of ab+cd

Because $a^2 + b^2 = 1$ we can take $a = \sin\alpha$ and $b = \cos \alpha $

Similarly $c=\sin \beta$ and $d = \cos\beta$

$ac + bd = \sin\alpha \sin  \beta + \cos \alpha \cos \beta = 0$

or $\cos (\alpha - \beta) = 0\cdots (1)$  (using formula of cos of difference)

$ab + cd = \sin \alpha \cos \alpha + \sin \beta \cos \beta = \frac{1}{2}(\sin 2 \alpha + \sin  2 \beta)$ (using formula for sin of twice angle)  

$= \frac{1}{2} \sin (\alpha + \beta) \cos (\alpha - \beta)$ using sum of sines

$= \frac{1}{2} \sin (\alpha + \beta) * 0$ using (1)

= 0