<s>[INST] How do you solve this trig/geometry question? In a quadrilateral $ABCD$, if
$\sin\left(\frac{A+B}2\right)\cos\left(\frac{A-B}2\right) + \sin\left(\frac{C+D}2\right)\cos\left(\frac{C-D}2\right) = 2$
then $\sin\left(\frac A 2\right) \sin\left(\frac B 2\right) \sin\left(\frac C 2\right) \sin\left(\frac D 2\right) = $
a) $\frac 1 8 $
b) $\frac1 4$ 
c) $\frac {1}{2\sqrt2}$ 
d) $\frac 1 2$
 [/INST] For $w$, $x$, $y$, $z$ between $0$ and $\pi$, the relation
$$\sin w \;\cos x \;+\; \sin y \; \cos z \;=\; 2$$
holds if and only if each term ---and each factor of each term--- is itself $1$. Specifically,
$$w = y = \frac{\pi}{2} \quad\text{ and }\quad x = z = 0$$

Now, note that any two angles of a quadrilateral have a sum and (absolute) difference between $0$ and $2\pi$ (because the sum of all angles is $2\pi$); thus, they have a half-sum and half-difference between $0$ and $\pi$.
Therefore, if for $\square ABCD$ we define
$$w := \frac{A+B}{2} \quad x := \frac{|A-B|}{2} \quad y := \frac{C+D}{2} \quad z := \frac{|C-D|}{2}$$ 
then the above applies to $w$, $x$, $y$, $z$ and allows us to conclude that $A=B=C=D = \pi/2$, so that
$$\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\sin\frac{D}{2} = \sin^4\frac{\pi}{4} =\left(\frac{\sqrt{2}}{2}\right)^4 = \frac{1}{4}$$
</s>