Poner $x=x+\frac{\alpha}{2}-\frac{\alpha}{2}$ para conseguir $$\sin(x)=\sin(x+\frac{\alpha}{2})\cos(\frac{\alpha}{2})-\sin(\frac{\alpha}{2})\cos(x+\frac{\alpha}{2})$$ y poner $x+\alpha=x+\frac{\alpha}{2}+\frac{\alpha}{2}$ para conseguir $$\sin(x+\alpha)=\sin(x+\frac{\alpha}{2})\cos(\frac{\alpha}{2})+\sin(\frac{\alpha}{2})\cos(x+\frac{\alpha}{2})$$
Ahora con $u=\sin(x+\frac{\alpha}{2})$ que tenemos:
$$\frac{1}{\sin(x)}+\frac{1}{\sin(x+\alpha)}=2\cos(\frac{\alpha}{2})\frac{u}{u^2-(\sin(\frac{\alpha}{2}))^2}$$
Haz lo mismo con la otra expresión con $v=\sin(y+\frac{\alpha}{2})$ . Se consigue fácilmente $u=v$ o $uv=-(\sin(\frac{\alpha}{2}))^2$ y es fácil, utilizando $x+y+\alpha<\pi$ para terminar.