9. If the internal angles $A, B, C$ of $\triangle A B C$ satisfy $\sin A=\cos B=\tan C$, find the value of $\cos ^{3} A+\cos ^{2} A-\cos A$.
Obviously, $A$ is an obtuse angle, then $\sin \left(180^{\circ}-A\right)=\sin \left(90^{\circ}-B\right) \Rightarrow A-B=90^{\circ}$, thus $C=270^{\circ}-2 A \Rightarrow \sin A=\tan \left(270^{\circ}-2 A\right)=\cot 2 A \Rightarrow \sin A \tan 2 A=1$ $\Rightarrow 2 \sin A \tan A=1-\tan ^{2} A \Rightarrow \frac{2 \sin ^{2} A}{\cos A}=1-\frac{\sin ^{2} A}{\cos ^{2} A}=\frac{\cos 2 A}{\cos ^{2} A}$
$\Rightarrow 2-2 \cos ^{2} A=\frac{2 \cos ^{2} A-1}{\cos A} \Rightarrow 2 \cos A-2 \cos ^{3} A=2 \cos ^{2} A-1$
$\Rightarrow 2 \cos ^{3} A+2 \cos ^{2} A-2 \cos A=1 \Rightarrow \cos ^{3} A+\cos ^{2} A-\cos A=\frac{1}{2}$.