1Or finish it easily.
Using formula sin2A+sin2B+sin2C=4sinAsinBsinC ,
We get sin2A+sin2B+sin2C2sinAsinBsinC
= 4sinAsinBsinC2sinAsinBsinC
= 2.
Anyways,thank you for the proof.
If A+B+C = π (pi radian)
Prove that:
cosAsinBsinC+cosBsinCsinA+cosCsinAsinB=2
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5 Answers
30LHS = cosAsinBsinC+cosBsinCsinA + cosCsinAsinB
= cosAsinA + cosBsinB + cosCsinCsinAsinBsinC
= sin2A + sin2B + sin2C2sinAsinBsinC
= 2sin(A+B)cos(A-B) + 2sinCcosC{cos(A-B) - cos(A+B)}sinC
now since, A+B+C=Ï€,
= 2sinCcos(A-B) + 2sinCcosC{cos(A-B) - cos(A+B)}sinC
= 2sinC{cos(A-B) + cosC}{cos(A-B) - (-cosC)}sinC
= 2 = RHS
1
341\sum \frac{\cos A}{\sin B \sin C} = - \sum \frac{\cos (B+C)}{\sin B \sin C}= 3 - \sum \cot A \cot B = 3-1=2
(Since in a triangle ABC, \sum \cot A \cot B = 1)