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prove sin²A - sin²B/sinA .cosA- sinB. cosB = tan(A+B)

$\frac{\sin^{2} A - \sin^{2} B}{\sin A cos A - \sin B cos B} = \frac{\frac{1 - \cos 2 A}{2} - \frac{1 - \cos 2 B}{2}}{\frac{2 \sin A \cos A}{2} - \frac{2 \sin B cos B}{2}} = \frac{1 - \cos 2 A - 1 + \cos 2 B}{\sin 2 A - \sin 2 B} = \frac{- (\cos 2 B - \cos 2 A)}{\sin 2 A - \sin 2 B} = \frac{- (- 2 \sin (\frac{2 A + 2 B}{2}) \sin (\frac{2 A - 2 B}{2}))}{2 \sin (\frac{2 A - 2 B}{2}) \cos (\frac{2 A+ 2 B}{2})} = \frac{\sin (A + B)}{\cos (A + B)} = \tan (A + B)$

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