Least value of $ab-cd$ is 
If $a,b,c,d\in\mathbb{R}$ and $a^2+b^2=c^2+d^2=4$
$ac+bd=0$. Then least value of $ab-cd$ is

Plan
$(ac+bd)^2+(ab-cd)^2=a^2c^2+b^2d^2+a^2b^2+c^2d^2\geq 0$
$a^2(b^2+c^2)+d^2(b^2+c^2)\geq 0$
How do i solve it Help me please
 A: Let $a=2\sin\alpha$, $b=2\cos\alpha$, $c=2\sin\beta$ and $d=2\cos\beta$.
Then $4\sin\alpha\sin\beta+4\cos\alpha\cos\beta=0$ and hence $\cos(\alpha-\beta)=0$. Therefore, $\alpha=(n+\frac12)\pi+\beta$.
\begin{align*}
ab-cd&=2\sin2\alpha-2\sin2\beta\\
&=4\cos(\alpha+\beta)\sin(\alpha-\beta)\\
&=4\cos\left(\left(n+\frac12\right)\pi+2\beta\right)\sin\left(\left(n+\frac12\right)\pi\right)
\end{align*}
The least possible value is $-4$.
A: Geometric interpretation of conditions above with $U=\binom{a}{b}$, $V=\binom{c}{d}$ give :
$\|U\|=\|V\|=2$ and dot product $U . V=0$, i.e., $U$ and $V$ are orthogonal. 
Thus $a = 2\cos \alpha, b=2\sin \alpha, c = 2\cos \beta, d=2\sin \beta \tag{1}$
for a certain $\alpha$ ($-\pi < \alpha \leq \pi$) with $\beta=\alpha\pm \pi/2,$ expressing orthogonality.
As $$ab-cd=4 \sin \alpha \cos \alpha-4 \sin  \beta \cos \beta$$
$$ab-cd=2 \sin 2 \alpha-2 \sin 2 \beta $$
with $2 \beta=2 \alpha\pm \pi$.
$$ab-cd=2 \sin 2 \alpha-2 \sin  (2 \alpha\pm \pi)$$
$$ab-cd=2 \sin 2 \alpha+2 \sin 2 \alpha=4 \sin 2 \alpha \tag{2}$$
with minimal value $-4$ for $\alpha=-\pi/4$, i.e., using (1) with $\beta=\alpha+\pi/2=\pi/4$ with :
$$a=\sqrt{2}, b=- \sqrt{2}, c= \sqrt{2}, d=\sqrt{2}.$$
More generaly with :
$$a=s_1 \sqrt{2}, b=- s_1 \sqrt{2}, c= s_2 \sqrt{2}, d=s_2 \sqrt{2} \ \text{with} \ s_1, s_2 = \pm 1$$
