$\begin{array}{l}
x \in \left( {0;\dfrac{\pi }{2}} \right) \to 0 < \sin \alpha < 1\\
\to {x^2} - 2x\sin \alpha + 1 > 0\forall x \in R\\
y = \dfrac{{{x^2}\sin \alpha - 2x + \sin \alpha }}{{{x^2} - 2x\sin \alpha + 1}}\\
y = \dfrac{{{x^2}\sin \alpha - 2x + \sin \alpha }}{{{x^2} - 2x\sin \alpha + 1}} + 1 - 1\\
= \dfrac{{\left( {1 + \sin \alpha } \right){x^2} - x\left( {2 + 2\sin \alpha } \right) + \left( {1 + \sin a} \right)}}{{{x^2} - 2x\sin \alpha + 1}} - 1\\
= \dfrac{{\left( {1 + \sin \alpha } \right)\left( {{x^2} - 2x + 1} \right)}}{{{x^2} - 2x\sin \alpha + 1}} - 1\\
= \dfrac{{\left( {1 + \sin \alpha } \right){{\left( {x - 1} \right)}^2}}}{{{x^2} - 2x\sin \alpha + 1}} - 1 \ge - 1\\
\Rightarrow \min y = - 1\\
y = \dfrac{{{x^2}\sin \alpha - 2x + \sin \alpha }}{{{x^2} - 2x\sin \alpha + 1}} - 1 + 1\\
y = \dfrac{{\left( {\sin \alpha - 1} \right){x^2} - 2x\left( {1 - \sin \alpha } \right) - \left( {1 - \sin \alpha } \right)}}{{{x^2} - 2x\sin \alpha + 1}} + 1\\
y = \dfrac{{\left( {1 - \sin \alpha } \right)\left( { - {x^2} - 2x - 1} \right)}}{{{x^2} - 2x\sin \alpha + 1}} + 1\\
y = \dfrac{{ - \left( {1 - \sin \alpha } \right){{\left( {x + 1} \right)}^2}}}{{{x^2} - 2x\sin \alpha + 1}} + 1 \le 1\\
\Rightarrow \max y = 1\\
\Rightarrow \left\{ \begin{array}{l}
M = 1\\
m = - 1
\end{array} \right. \Rightarrow M + m = 0 \to A
\end{array}$
