Let A, B, C are three angles such that sinA + sinB + sinC = 0, then frac {sinAsin BsinC}{(sin 3A+ si

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3.Trigonometrical Ratios, Functions and Identities

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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

$\frac{1}{12}$

Solution

$\frac{\sin A \sin B \sin C}{\left(3(\sin A+\sin B+\sin C)-4\left(\sin ^{3} A+\sin ^{3} B+\sin ^{3} C\right)\right)}$

$=-\frac{1}{12}$

Standard 11

Mathematics

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