This is equivalent to \begin{align} |\alpha \sin A + \sin(A+B)|&<|\sin B|\\ ((\alpha+\cos B) \sin A + \cos A \sin B)^2&<(\sin B)^2\\ ((\alpha + \cos B)^2-\sin^2 B)\sin^2 A &<-2(\alpha+\cos\beta)\sin A \cos A \sin B\\ \frac{(\alpha + \cos B)^2-\sin^2 B}{\sin B} &<\frac{-2(\alpha+\cos\beta)\cos A}{\sin A}\\ \end{align}
So the area in question is the sum of the areas with $$\frac{\sin^2 B-(\alpha + \cos B)^2}{|\alpha + \cos B|\sin B} >2\cot A,\ \ \alpha+\cos B > 0$$ $$\frac{\sin^2 B-(\alpha + \cos B)^2}{|\alpha + \cos B|\sin B} >-2\cot A,\ \ \alpha+\cos B < 0$$
Since $-\cot A=\cot(\pi-A)$, the area in question equals the area with $$\frac{\sin^2 B-(\alpha + \cos B)^2}{|\alpha + \cos B|\sin B} >2\cot A$$ This is equivalent to $$1/\frac{|\alpha + \cos B|}{\sin B}-\frac{|\alpha + \cos B|}{\sin B} >1/\tan\left(\frac{A}{2}\right)-\tan\left(\frac{A}{2}\right)$$ and therefore to $$\frac{|\alpha + \cos B|}{\sin B} <\tan\left(\frac{A}{2}\right)$$ So the area in question can also be written as $$A(\alpha)=\int_{B=0}^{\pi} 2\arctan\frac{|\alpha + \cos B|}{\sin B} dB$$$${\cal A}(\alpha)=\int_{B=0}^{\pi} 2\arctan\frac{|\alpha + \cos B|}{\sin B} dB$$ Now it is easy to verify $A(0)=\pi^2/2$${\cal A}(0)=\pi^2/2$, and \begin{align} \frac{dA(\alpha)}{d\alpha}&=\int_{0}^{\arccos(-\alpha)}\frac{2 \sin B\, dB}{1+2\alpha \cos B+\alpha^2}dB- \int_{\arccos(-\alpha)}^\pi\frac{2 \sin B\, dB}{1+2\alpha \cos B+\alpha^2}\\ &=\frac{1}{2\alpha}\log\frac{1+\alpha}{1-\alpha}-\frac{1}{2\alpha}\log\frac{1+\alpha}{1-\alpha}\\ &=0 \end{align}\begin{align} \frac{d{\cal A}(\alpha)}{d\alpha}&=\int_{0}^{\arccos(-\alpha)}\frac{2 \sin B\, dB}{1+2\alpha \cos B+\alpha^2}dB- \int_{\arccos(-\alpha)}^\pi\frac{2 \sin B\, dB}{1+2\alpha \cos B+\alpha^2}\\ &=\frac{1}{2\alpha}\log\frac{1+\alpha}{1-\alpha}-\frac{1}{2\alpha}\log\frac{1+\alpha}{1-\alpha}\\ &=0 \end{align} which leads to $A(\alpha)=\pi^2/2$${\cal A}(\alpha)=\pi^2/2$ for all $\alpha$.
