If z(1), z(2) and z(3) are the vertices ...

If `z_(1)`, `z_(2)` and `z_(3)` are the vertices of `DeltaABC`, which is not right angled triangle taken in anti-clock wise direction and `z_(0)` is the circumcentre, then `((z_(0)-z_(1))/(z_(0)-z_(2)))(sin2A)/(sin2B)+((z_(0)-z_(3))/(z_(0)-z_(2)))(sin2C)/(sin2B)` is equal to

`0`

`1`

`-1`

`2`

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Verified by Experts

The correct Answer is:

`(c )` Takin rotation at `'O'` ltbr. `(z_(0)-z_(1))/(z_(0)-z_(2))=cos2C-isin2C`
`(z_(0)-z_(3))/(z_(0)-z_(2))=cos2A+isin2A`
Now `((z_(0)-z_(1))/(z_(0)-z_(2)))(sin2A)/(sin2B)+((z_(0)-z_(3))/(z_(0)-z_(2)))(sin2C)/(sin2B)`
`=sin2Acos2C-isin2Asin2C`
`=(+cos2Asin2C+isin2Asin2C)/(sin2B)`
`=(sin(2A+2C))/(sin2B)=-1`

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