The range of $\cos \theta \left(\sin \theta +\sqrt{{\sin }^2\theta +{\sin }^2\alpha }\right)$  is $\left(a,b\right)$ , then

$y\sec \theta =\sin \theta +\sqrt{{\sin }^2\theta +{\sin }^2\alpha }$

${\left(y\sec \theta -\sin \theta \right)}^2={\left(\sqrt{{\sin }^2\theta +{\sin }^2\alpha }\right)}^2$

$y^2{\sec }^2\theta +{\sin }^2\theta -2y\sec \theta \sin \theta ={\sin }^2\theta +{\sin }^2\alpha$

$y^2\left(1+{\tan }^2\theta \right)-2y\tan \theta ={\sin }^2\alpha$

$y^2+y^2{\tan }^2\theta -2y\tan \theta ={\sin }^2\alpha$

$y^2{\tan }^2\theta -2y\tan \theta +\left(y^2-{\sin }^2\alpha \right)=0$

$\tan \theta \in R$

$D\ge 0$

$b^2-4ac\ge 0$

${\left(-2y\right)}^2-4y^2\left(y^2-{\sin }^2\alpha \right)\ge 0$

$4y^2-4y^4+4y^2{\sin }^2\alpha \ge 0$

$4y^2\left[1-y^2+{\sin }^2\alpha \right]>0$

$4y^2\ge 01-y^2+{\sin }^2\alpha \ge 0$

$y^2\le 1+{\sin }^2\alpha$

$\left|y\right|\le \left(\sqrt{1+{\sin }^2\alpha }\right)$

$-\sqrt{1+{\sin }^2\alpha }\le y\le +\sqrt{1+{\sin }^2\alpha }$

$\left[-\sqrt{1+{\sin }^2\alpha },\sqrt{1+{\sin }^2\alpha }\right]$