જો A, B, C એ ત્રણ ખૂણા છે કે જેથી sinA + sinB + sinC = 0

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3.Trigonometrical Ratios, Functions and Identities

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જો $A, B, C$ એ ત્રણ ખૂણા છે કે જેથી $sinA + sinB + sinC = 0,$ થાય તો

$ \frac {sinAsin BsinC}{(sin 3A+ sin 3B+ sin 3C)}$ (wherever definied)=

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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