Given equation is
$$
x^2-(\sin \alpha-2) x-(1+\sin \alpha)=0
$$
Let $x_1$ and $x_2$ be two roots of quadratic equation.
$$
\begin{aligned}
& \therefore x_1+x_2=\sin \alpha-2 \text { and } x_1 x_2=-(1+\sin \alpha) \\
& \left(x_1+x_2\right)^2=(\sin \alpha-2)^2=\sin ^2 \alpha+4-4 \sin \alpha
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad x_1^2+x_2^2=\sin ^2 \alpha+4-4 \sin \alpha-2 x_1 x_2 \\
& =\sin ^2 \alpha+4-4 \sin \alpha+2(1+\sin \alpha) \\
& =\sin ^2 \alpha-2 \sin \alpha+6
\end{aligned}
$$
Now, By putting
$\alpha=\frac{\pi}{6}, \alpha=\frac{\pi}{4}, \alpha=\frac{\pi}{3}$ and $\alpha=\frac{\pi}{2}$ in (A) one by one
We get least value of $x_1^2+x_2^2$ at $\frac{\pi}{2}$
Hence, $\alpha=\frac{\pi}{2}$