  The principal aim of this paper is to employ Bessel-type operators in proving
the inequality \begin{align*} \int_0^\pi dx \, |f'(x)|^2 \geq
\dfrac{1}{4}\int_0^\pi dx \, \dfrac{|f(x)|^2}{\sin^2
(x)}+\dfrac{1}{4}\int_0^\pi dx \, |f(x)|^2,\quad f\in H_0^1 ((0,\pi)),
\end{align*} where both constants $1/4$ appearing in the above inequality are
optimal. In addition, this inequality is strict in the sense that equality
holds if and only if $f \equiv 0$. This inequality is derived with the help of
the exactly solvable, strongly singular, Dirichlet-type Schr\"{o}dinger
operator associated with the differential expression \begin{align*}
\tau_s=-\dfrac{d^2}{dx^2}+\dfrac{s^2-(1/4)}{\sin^2 (x)}, \quad s \in
[0,\infty), \; x \in (0,\pi). \end{align*}
  The new inequality represents a refinement of Hardy's classical inequality
\begin{align*} \int_0^\pi dx \, |f'(x)|^2 \geq \dfrac{1}{4}\int_0^\pi dx \,
\dfrac{|f(x)|^2}{x^2}, \quad f\in H_0^1 ((0,\pi)), \end{align*} it also
improves upon one of its well-known extensions in the form \begin{align*}
\int_0^\pi dx \, |f'(x)|^2 \geq \dfrac{1}{4}\int_0^\pi dx \,
\dfrac{|f(x)|^2}{d_{(0,\pi)}(x)^2}, \quad f\in H_0^1 ((0,\pi)), \end{align*}
where $d_{(0,\pi)}(x)$ represents the distance from $x \in (0,\pi)$ to the
boundary $\{0,\pi\}$ of $(0,\pi)$.
