Prove tan (A + B) = (sin^2 A - sin^2 B) / (sin A cos A - sin B cos B)

Trigonometric Identity

we have sin A cos A - sin B cos B = 1/2( sin 2A - sin 2B) = 1/2 * 2 cos (A+B)cos (A-B)
using sin C -sin D = 2 cos (C +D)/2 cos (C -D)/2

so sin A cos A - sin B cos B = cos (A+B)cos (A-B) ...1

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

so I get tan (A + B) tan (A- B) = (sin^2 A - sin^2 B) / (sin A cos A - sin B cos B) and not what you have specified