<s>[INST] I am trying to prove that, ${\sin{A}\cos{A}-\sin{B}\cos{B}\over \sin^2{A}-\sin^2{B}}=\tan(90^o-A-B)$ I am trying to prove that,
$${\sin{A}\cos{A}-\sin{B}\cos{B}\over \sin^2{A}-\sin^2{B}}=\tan(90^o-A-B)$$
Using: $\sin(2A)=2\sin{A}\cos{A}$, then we have
$${1\over 2}{\sin(2A)-\sin(2B)\over \sin^2{A}-\sin^2{B}}=\tan(90^0-A-B)$$
$${1\over 2}{\sin(2A)-\sin(2B)\over (\sin{A}-\sin{B})(\sin{A}+\sin{B})}=\tan(90^0-A-B)$$
I am out of ideas, I need some help, anyone's?
 [/INST] $$
\frac{\sin A \cos A - \sin B \cos B}{\sin^2 A - \sin^2 B} = \frac{\sin (x-y) \cos (x+y)}{\sin (x-y) \sin(x+y)} =
$$
$$
= \frac{\cos (x+y) }{\sin (x+y)}  = \cot (x+y) = \tan (90 - x-y)
$$
</s>