$A= \dfrac{\sin(90^\circ - \alpha) + \cos(180^\circ - \alpha)}{\tan \alpha \cdot \cot(90^\circ - \alpha)}$

$= \dfrac{\cos \alpha + (-\cos \alpha)}{\tan \alpha \cdot \tan \alpha} $

$= \dfrac{\cos \alpha - \cos \alpha}{\tan^2 \alpha}$

$= \dfrac{0}{\tan^2 \alpha} = 0$

Vậy `A=0`