To prove: $\bigl(1 + \sin(A+B)\bigr)\bigl(1-\sin(A-B)\bigr) = (\cos A - \sin B)^2 $ How could one show that
$$ \bigl(1 + \sin(A+B)\bigr)\bigl(1-\sin(A-B)\bigr) = (\cos A - \sin B)^2
$$
I have tried the below approach:
\begin{align*}
LHS & = \bigl(1 + \sin(A+B)\bigr)\bigl(1-\sin(A-B)\bigr)\\
& =  1-\sin(A-B)+\sin(A+B)-\sin(A+B)\sin(A-B)\\
& = 1 - \sin{A}\cos{B} + \sin{B}\cos{A} + \sin{A}\cos{B} + \sin{B}\cos{A}\\
& \qquad - (\sin{A})^2 + (\sin{B})^2\\
& = 1 + 2\sin{B}\cos{A} - 1 + (\cos{A})^2 + (\sin{B})^2\\
& = 2\sin{B}\cos{A} + (\cos{A})^2 + (\sin{B})^2\\
& = (\cos{A} + \sin{B})^2
\end{align*}

$$(1+\sin A\cos B+\cos A\sin B)(1-\sin A\cos B+\cos A\sin B)$$
$$=(1+\cos A\sin B)^2-(\sin A\cos B)^2$$
$$=1+2\cos A\sin B+\cos^2A\sin^2B-(1-\cos^2A)(1-\sin^2B)$$
$$=\cdots$$
$$=(1+\cos A\sin B)^2$$
