Question

In a quadrilateral, 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 $\sum \cos \frac{A}{2}.\cos \frac{B}{2}$ is equal to

• A. $0$
• B. $6$
• C. $3$
• D. $2$

Answer 1

Answer: (C)

$3$

We know, in a quadrilateral, $A+B+C+D=2\pi$

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

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

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

$\Rightarrow 2\text{sin}\left(\frac{A+B}{2}\right)\text{cos}\left(\frac{A-B+C-D}{4}\right)\text{cos}\left(\frac{A-B-C+D}{4}\right)=2$

So,

$\text{sin}\left(\frac{A+B}{2}\right)=1\Rightarrow A+B=\pi ,C+D=\pi \to \left(1\right)$

$\text{cos}\left(\frac{A-B+C-D}{4}\right)=1\Rightarrow A+C=B+D\to \left(2\right)$

$\text{cos}\left(\frac{A-B-C+D}{4}\right)=1\Rightarrow A+D=B+C\to \left(3\right)$

Adding $\left(2\right)$ and $\left(3\right),$ we get$,$ $A=B$

Subtracting $\left(2\right)$ and $\left(3\right),$ we get$,$ $C=D$

Using $\left(1\right),$ we can conclude, $A=B=C=D=\frac{\pi }{2}$

So, $\sum \text{cos}\frac{A}{2}\text{cos}\frac{B}{2}=6\times \frac{1}{2}=3$