A3. Let $\alpha$ and $\beta$ be the acute angles of a right triangle. Which of the following expressions is equivalent to the expression $\frac{\sin \left(90^{\circ}-\alpha\right)+\cos \beta}{4 \sin \beta \cdot \cot \left(90^{\circ}-\alpha\right)}$?
(A) $\frac{\sin \beta+\cos \beta}{4 \cos \beta}$
(B) $\frac{\sin \beta+\cos \beta}{4 \sin \beta}$
(C) $\frac{1}{4 \cos \beta}$
(D) $\frac{\sin \beta-\cos \beta}{4 \cos \beta}$
(E) $\frac{\sin \beta+\cos \beta}{2 \cos \beta}$
A3. We use the relations $\sin \left(90^{\circ}-\alpha\right)=\cos \beta, \cot \left(90^{\circ}-\alpha\right)=\frac{\cos \left(90^{\circ}-\alpha\right)}{\sin \left(90^{\circ}-\alpha\right)}=\frac{\cos \beta}{\sin \beta}$. The original expression is transformed into $\frac{(\sin \beta+\cos \beta) \cdot \sin \beta}{4 \sin \beta \cdot \cos \beta}$, we simplify and get the result $\frac{\sin \beta+\cos \beta}{4 \cos \beta}$.