Question 50, If the internal angles $A, B, C$ of $\triangle ABC$ satisfy $\sin A=\cos B=\tan C$, find the value of $\cos ^{3} A+\cos ^{2} A-$ $\cos A$.
Question 50, Solution: Since $\sin A=\cos B$, and $\tan C=\sin A \in(0,1]$, hence $A=B+90^{\circ}, C=$ $270^{\circ}-2 A$. Therefore, according to the conditions:
$$
\begin{array}{l}
\sin A=\tan C=\tan \left(270^{\circ}-2 A\right)=\cot 2 A=\frac{\cos 2 A}{\sin 2 A}=\frac{2 \cos ^{2} A-1}{2 \sin A \cos A} \\
\Rightarrow 2 \cos ^{2} A-1=2 \cos A \sin ^{2} A=2 \cos A\left(1-\cos ^{2} A\right) \\
\Rightarrow \cos ^{3} A+\cos ^{2} A-\cos A=\frac{1}{2}
\end{array}
$$

Therefore, the required $\cos ^{3} A+\cos ^{2} A-\cos A=\frac{1}{2}$.