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Question

cosAsinBsinC+cosBsinCsinA+cosCsinAsinB=2.

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Solution

On simplification, we have to prove that sinAcosA+sinBcosB+sinCcosCsinAsinBsinC=2.
Multiplying both sides by 2, we get
sin2A+sin2B+sin2C=4sinAsinBsinC.

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