1. Given that $\alpha$ is an acute angle of a right triangle. Then $\sqrt{\left(1-\frac{\tan \alpha}{\sin \alpha}\right)^{2}}=(\quad)$.
(A) $1-\frac{\tan \alpha}{\sin \alpha}$
(B) $\frac{\tan \alpha}{\sin \alpha}-1$
(C) $1+\frac{\tan \alpha}{\sin \alpha}$
(D) Cannot be determined
$-1 . B$
From the definition of trigonometric functions, we get $\tan \alpha>\sin \alpha$.
Then $\sqrt{\left(1-\frac{\tan \alpha}{\sin \alpha}\right)^{2}}=\frac{\tan \alpha}{\sin \alpha}-1$.