Problem (Algebra):
$\quad$ Compute the least positive value of $t$ such that

$$
\operatorname{Arcsin}(\sin (t)), \operatorname{Arccos}(\cos (t)), \operatorname{Arctan}(\tan (t))
$$

form (in some order) a three-term arithmetic progression with a nonzero common difference.

Solution:
For $0 \leq t<\pi / 2$, all three values are $t$, so the desired $t$ does not lie in this interval.

For $\pi / 2<t<\pi$,

$$
\begin{aligned}
\operatorname{Arcsin}(\sin (t)) & =\pi-t \in(0, \pi / 2) \\
\operatorname{Arccos}(\cos (t)) & =t \quad \in(\pi / 2, \pi) \\
\operatorname{Arctan}(\tan (t)) & =t-\pi \in(-\pi / 2,0)
\end{aligned}
$$

A graph of all three functions is shown below.



Thus if the three numbers are to form an arithmetic progression, they should satisfy

$$
t-\pi<\pi-t<t
$$

The three numbers will be in arithmetic progression if and only if $t+(t-\pi)=2(\pi-t)$, which implies $t=\frac{\mathbf{3 \pi}}{\mathbf{4}}$. Note that if $t=\frac{3 \pi}{4}$, the arithmetic progression is $-\frac{\pi}{4}, \frac{\pi}{4}, \frac{3 \pi}{4}$, as required.

Final answer: $\frac{3 \pi}{4}$