Ta có
$E = \dfrac{2\sin(a+b)}{\cos(a + b) + \cos(a-b)} - \tan b$
$= \dfrac{2(\sin a \cos b + \sin b \cos a}{2 \cos a \cos b} - \tan b$
$= \dfrac{\sin a \cos b}{\cos a \cos b} + \dfrac{\sin b \cos a}{\cos a \cos b} - \tan b$
$= \dfrac{\sin a}{\cos a} + \dfrac{\sin b}{\cos b} - \tan b$
$= \tan a + \tan b - \tan b = \tan a$.