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*}
 A: \begin{aligned}
\text { LHS }=&(1+\sin (A+B))(1-\sin (A-B)) \\
=& 1-\sin (A-B)+\sin (A+B)-\sin (A+B) \sin (A-B) \\
=& 1- \sin A \cos B + \cos A \sin B + \sin A \cos B + \cos A \sin B -\sin (A+B) \sin (A-B)\\
=& 1 + 2\cos A \sin B - \sin (A+B) \sin (A-B) \\
=& 1 + 2\cos A \sin B - \frac{1}{2} (cos(A+B-A+B) - cos(A+B+A-B)) \\
=& 1 + 2\cos A \sin B - \frac{1}{2} (cos(2B) - cos(2A)) \\
=& 1 + 2\cos A \sin B - \frac{1}{2} [(1 - 2 \sin^{2} B )-(2 \cos^{2} A - 1)] \\
=& 1 + 2\cos A \sin B - \frac{1}{2} (1 - 2 \sin^{2} B ) +\frac{1}{2} (2 \cos^{2} A - 1) \\
=&  2\cos A \sin B + \sin^{2} B + \cos^{2} A\\
=& \cos^{2} A - 2\cos A \sin B + \sin^{2} B \\
=&(\cos A + \sin B)^{2}
\end{aligned}
Maybe the exercise is really wrong.
I am attaching a link where you can see the formulas used. (https://doza.pro/art/math/geometry/en/trig-formulas)
I hope I helped you.
A: $$(1+\sin A\cos B+\cos A\sin B)(1-\sin A\cos B+\cos A\sin B)$$
$$=(1+\cos A\sin B)^2-(\sin A\cos B)^2$$
$$=1+2\cos A\sin B+\cos^2A\sin^2B-(1-\cos^2A)(1-\sin^2B)$$
$$=\cdots$$
$$=(1+\cos A\sin B)^2$$
