24 In triangle $A B C$, $\angle C>90^{\circ}$, let $x=\tan A \tan B, y=\cot A \cot B+\cot B \cot C+$ $\cot C \cot A, z=\sin A-\cos (B+C)$, then
A. $x<y<z$
B. $y<x<z$
C. $z<y<x$
D. $x<z<y$
24 A.
$$
\begin{array}{c}
y=\cot A \cot B-\cot (A+B)(\cot A+\cot B) \\
=\cot A \cot B-(\cot A \cot B-1)=1 . \\
z=\sin A+\cos A>1=y .
\end{array}
$$

Since $C>90^{\circ}$, we have $\tan A>0, \tan B>0, \tan (A+B)>0$, so $\frac{\tan A+\tan B}{1-\tan A \tan B}$
$$
>0, x=\tan A \tan B<1 .
$$