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?
 A: $$
\frac{\sin A \cos A - \sin B \cos B}{\sin^2 A - \sin^2 B} = \frac{\sin (x-y) \cos (x+y)}{\sin (x-y) \sin(x+y)} =
$$
$$
= \frac{\cos (x+y) }{\sin (x+y)}  = \cot (x+y) = \tan (90 - x-y)
$$
A: From where you set off:
$$ {1\over 2}{\sin(2A)-\sin(2B)\over \sin^2{A}-\sin^2{B}} \ = \ {1\over 2} \ \cdot \ {\frac{\sin(2A)-\sin(2B)}{[ \frac{1}{2} ( \ 1 - \cos[2A]) \ ] \ - \  [ \frac{1}{2} ( \ 1 - \cos[2B] \ ) \ ]}}$$
[applying "sine-squared identity"]
$$ = \ {1\over 2} \ \cdot \ {\frac{\sin(2A)-\sin(2B)}{  \frac{1}{2} (  \ \cos[2B] - \cos[2A] \ ) \    }} \ = \ {\frac{\sin(2A)-\sin(2B)}{      \ \cos(2B) - \cos(2A)      }} $$
$$ = \ {\frac{2 \cos \left( \frac{2A+2B}{2} \right) \sin \left( \frac{2A-2B}{2} \right)}{ -  2 \sin \left( \frac{2B+2A}{2} \right) \sin \left( \frac{2B-2A}{2} \right)    }} \ = \ {\frac{  \cos  ( A + B) \sin \left( \frac{2A-2B}{2} \right)}{     \sin  ( A + B) \sin \left( \frac{2A-2B}{2} \right)    }} \ = \ {\frac{  \cos  ( A + B)  }{     \sin  ( A + B)      }} $$
[applying "sum-to-product" identities and odd symmetry property of sine function]
$$= \   \cot  ( A + B) \ = \  \tan( \ 90º \ - \ [A+B] \ ) \ \ . $$ 
[This is not greatly different from Olba12's derivation, other than to proceed along the path you'd started on.  There isn't really any getting around having to use either the product-to-sum or sum-to-product identities.]
EDIT (later that day) --  Here's a way to show this starting from the right-hand side:
$$ \tan( \ 90º \ - \ A \ - \ B \ ) \ = \     \cot  ( A + B) \ = \ \frac{1}{ \tan  ( A + B)} \ = \ \frac{1 \ - \ \tan A \ \tan B}{\tan A \ + \ \tan B}    $$ 
$$ = \ \frac{1 \ - \ \tan A \ \tan B}{\tan A \ + \ \tan B} \ \cdot \ \frac{\tan A \ - \ \tan B}{\tan A \ - \ \tan B}  \ = \ \frac{\tan A \ - \ \tan B \ - \ \tan^2 A \ \tan B \ + \ \tan A \ \tan^2 B}{\tan^2 A \ - \ \tan^2 B}  $$ 
$$= \ \frac{(1 \ + \ \tan^2 B) \ \tan A \ - \ (1 \ + \ \tan^2 A) \ \tan B }{\tan^2 A \ - \ \tan^2 B} \ = \ \frac{\sec^2 B \ \tan A \ - \ \sec^2 A \ \tan B }{\tan^2 A \ - \ \tan^2 B}  $$ 
$$ = \ \frac{\frac{\sin A}{\cos A \cos^2 B} \ - \ \frac{\sin B}{\cos B \cos^2 A} }{\frac{\sin^2 A}{\cos^2 A} \ - \ \frac{\sin^2 B}{\cos^2 B}} \ = \ \frac{ \frac{\sin A \cos A \ - \ \sin B \cos B}{\cos^2 A \cos^2 B} }{\frac{\sin^2 A \cos^2 B \ - \ \sin^2 B \cos^2 A}{\cos^2 A \cos^2 B}} $$
$$ = \  \frac{\sin A \cos A \ - \ \sin B \cos B}{\sin^2 A \cos^2 B \ - \ \sin^2 B \cos^2 A} \ = \ \frac{\sin A \cos A \ - \ \sin B \cos B}{\sin^2 A \ ( 1 - \sin^2 B) \ - \ \sin^2 B \ ( 1 - \sin^2 A)} $$
$$ = \ \frac{\sin A \cos A \ - \ \sin B \cos B}{\sin^2 A \   - \ \sin^2 A \sin^2 B \ - \ \sin^2 B \   + \  \sin^2 B \sin^2 A } \ = \ \frac{\sin A \cos A \ - \ \sin B \cos B}{\sin^2 A \     - \ \sin^2 B   } \ \ . $$
All right, there is a way to do it without the more advanced identities -- it just requires rather more writing...
