If $A (z_{1}), B (z_{2}) and C (z_{3})$ be the vertices of triangle ABC, taken in anti-clock wise direction and $z_{0}$ be the circumcentre, then $(\frac{z_{0} - z_{1}}{z_{0} - z_{2}}) \frac{\sin 2 A}{\sin 2 B} + (\frac{z_{0} - z_{3}}{z_{0} - z_{2}}) \frac{\sin 2 C}{\sin 2 B}$ is equal to

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answer is C.

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Detailed Solution

Taking rotation at O

$\frac{z_{0} - z_{1}}{z_{0} - z_{2}} = \cos 2 C - i \sin 2 C$

$\frac{z_{0} - z_{3}}{z_{0} - z_{2}} = \cos 2 A + i \sin 2 A$

Now $(\frac{z_{0} - z_{1}}{z_{0} - z_{2}}) \frac{\sin 2 A}{\sin 2 B} + (\frac{z_{0} - z_{3}}{z_{0} - z_{2}}) \frac{\sin 2 C}{\sin 2 B}$

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

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

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