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If cos alpha+cos beta+cos gamma=0=sin al...

If `cos alpha+cos beta+cos gamma=0=sin alpha+sin beta+sin gamma`, then `(sin3alpha+sin3beta+sin3gamma)/(sin(alpha+beta+gamma))` is equal to

A

`1`

B

`-1`

C

`3`

D

`-3`

Text Solution

Verified by Experts

The correct Answer is:
C
`(a)` `a=e^(ialpha)`, `b=e^(ibeta)`, `c=e^(igamma)`
Clearly `a+b+c=0`
`impliesa+b+c=0=(1)/(a)+(1)/(b)+(1)/(c )`
`impliesa^(3)+b^(3)+x^(3)=3abc`
`impliessin3alpha+sin3beta+sin3gamma=sin(alpha+beta+gamma)`
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