Let $A, B, C$ are three angles such that $sinA + sinB + sinC = 0,$ then
$ \frac {sinAsin BsinC}{(sin 3A+ sin 3B+ sin 3C)}$ (wherever definied) is -
Let $A, B, C$ are three angles such that $sinA + sinB + sinC = 0,$ then
$ \frac {sinAsin BsinC}{(sin 3A+ sin 3B+ sin 3C)}$ (wherever definied) is -
- A
$12$
- B
$-12$
- C
$ - \frac{1}{12}$
- D
$\frac{1}{12}$
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