Question

If A, B, C, D be the angles of a quadrilateral, what is the value of $\frac{\sin (A+B)}{\sin (S+D)}+\frac{\cos (C+D)}{\cos (A+B)}$?

Answer 1

$A+B+C+D={360}^0$
$\therefore A+B={360}^0-(C+D)$
or $\sin (A+B)=\sin [{360}^0-(C+D)]$
$=-\sin (C+D)$
and $\cos (A+B)=\cos [{360}^0-(C+D)]$
$\cos =(C+D)$
$\therefore$ the given expression =
$-\frac{\sin (C+D)}{\sin C+D}+\frac{\cos (C+D)}{\cos (C+D)}$
$=-1+1=0$