If $\text{A}\left({\text{z}}_{\text{1}}\right),\text{ B }\left({\text{z}}_{\text{2}}\right)\text{ and C }\left({\text{z}}_{\text{3}}\right)$ be the vertices of triangle ABC, taken in anti-clock wise direction and ${\text{z}}_0$  be the circumcentre, then $\left(\frac{z_0-z_1}{z_0-z_2}\right)\frac{\sin 2A}{\sin 2B}+\left(\frac{z_0-z_3}{z_0-z_2}\right)\frac{\sin 2C}{\sin 2B}$  is equal to

Taking rotation at O

$\frac{z_0-z_1}{z_0-z_2}=\cos 2C-i\sin 2C$

$\frac{z_0-z_3}{z_0-z_2}=\cos 2A+i\sin 2A$

Now      $\left(\frac{z_0-z_1}{z_0-z_2}\right)\frac{\sin 2A}{\sin 2B}+\left(\frac{z_0-z_3}{z_0-z_2}\right)\frac{\sin 2C}{\sin 2B}$

                                = $\frac{\text{sin2A cos2C - isin2A sin2C + cos2Asin2C + isin2Asin2C}}{\sin 2B}$

                               $=\frac{\sin (2A+2C)}{\sin 2B}=-1$