Question

If $\cos A+\sin B=m$ and $\sin A+\cos B=n,$ prove that $2\sin (A+B)=m^2+n^2-2.$

Answer 1

Given: $\cos A+\sin B=m$ and $\sin A+\cos B=n$
Now,
$2\sin (A+B)$
$=2(\sin A\cos B+\cos A\sin B)$
$=2+2\sin A\cos B+2\cos A\sin B-2$
$={\sin }^{2}A+{\cos }^{2}A+{\sin }^{2}B+{\cos }^{2}B$
$+2\sin A\cos B+2\cos A\sin B-2$
$={\sin }^{2}A+{\cos }^{2}B+2\sin A\cos B$
$+{\cos }^{2}A+{\sin }^{2}B+2\cos A\sin B-2$
$=(\sin A+\cos B{)}^2+(\cos A+\sin B{)}^2-2$
$=n^2+m^2-2$
Hence proved