(sin^(2)a-sin^(2)(b))/(sin(a)cos(a)-sin(b)cos(b))=tan(a+b)
Consider these identities:
$\displaystyle sin(a+b)sin(a-b)=sin^2a-sin^2b$
$\displaystyle sin(2a)=2sin(a)cos(a)$
$\displaystyle sin(2a)-sin(2b)=2cos(a+b)sin(a-b)$
Please try to prove these identities.
Using these your equation becomes:
$\displaystyle \frac{sin(a+b)sin(a-b)}{cos(a+b)sin(a-b)}=tan(a+b)$