$y\sec\theta = \sin\theta + \sqrt{\sin^2\theta+3}$, so $(y\sec\theta - \sin\theta)^2 = \sin^2\theta + 3$, giving $(y\tan\theta)^2 - 2y\tan\theta + y^2 - 3 = 0$. For real solutions in $\tan\theta$, discriminant $D = (2y)^2 - 4y^2(y^2-3) \geq 0 \Rightarrow 4 - 4(y^2-3) \geq 0 \Rightarrow y^2 \leq 4$, so $-2 \leq y \leq 2$.