show that (sin A + cos A) (Sin B + Cos B)= Sin (A +B) + Cos (A-B)
$\displaystyle RHS = \sin(A+B) + \cos(A-B) $
Expand both of these:
$\displaystyle =\sin(A)\cos(B)+\sin(B)\cos(A) + \cos(A)\cos(B) + \sin(A)\sin(B)$
Now factor out sin(A) and cos(A) from the relevant expressions:
$\displaystyle =\sin(A)(\cos(B)+\sin(B)) + \cos(A)(\sin(B) + \cos(B))$
Now realise that the inside of the brackets of both sides are identical. Hence the expression in the brackets is a factor of both terms. Take it our as a factor:
$\displaystyle =(\cos(B)+\sin(B))(\sin(A)+\cos(A)) = LHS $