Question 147
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) $\frac{\sin \beta+\cos \beta}{2 \cos \beta}$
$$
\frac{\sin \left(90^{\circ}-\alpha\right)+\cos \beta}{4 \sin \beta \cdot \cot \left(90^{\circ}-\alpha\right)}
$$
?
(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) $\frac{\sin \beta+\cos \beta}{2 \cos \beta}$