32. Let $A$ be an angle such that $\tan 8 A=\frac{\cos A-\sin A}{\cos A+\sin A}$. Suppose $A=x^{\circ}$ for some positive real number $x$. Find the smallest possible value of $x$.
32. Answer. 5
Since $\frac{\sin 8 A}{\cos 8 A}=\tan 8 A=\frac{\cos A-\sin A}{\cos A+\sin A}$, by rearranging we obtain
$$
\begin{array}{c}
\sin 8 A(\cos A+\sin A)=\cos 8 A(\cos A-\sin A) \\
\sin 8 A \cos A+\cos 8 A \sin A=\cos 8 A \cos A-\sin 8 A \sin A \\
\sin (8 A+A)=\cos (8 A+A) \\
\sin 9 A=\cos 9 A
\end{array}
$$
which reduces to
$$
\tan 9 A=1
$$

The smallest possible of $9 A=45^{\circ}$, which gives $x=5$.