Question

In 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 the value $\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\sin \frac{D}{2}$ is $\frac{1}{a}$ Find a

Answer 1

$\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)=2$

$\Rightarrow$$\frac{1}{2}[\sin A+\sin B+\sin C+\sin D]=2$

$\Rightarrow$$\sin A+\sin B+\sin C+\sin D=4$

$\therefore$ $\sin A=1,\sin B=1,\sin C=1,\sin D=1$

$\therefore$ $A+B+C+D={360}^0$......As it is a quadrilateral

$\therefore$ $A=B=C=D={90}^0$

$\Rightarrow$$\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\sin \frac{D}{2}=\frac{1}{4}$

Ans:a = 4