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Question

Prove: sin(AB)sinAsinB+sin(BC)sinBsinC+sin(CA)sinCsinA=0

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Solution

Given: sin(AB)sinAsinB+sin(BC)sinBsinC+sin(CA)sinCsinA=0

LHS: sin(AB)sinAsinB+sin(BC)sinBsinC+sin(CA)sinCsinA

We know that sin(α+β)=sinαcosβcosαsinβ

=sinAcosBcosAsinBsinAsinB+sinBcosCcosBsinCsinBsinC+sinCcosAcosCsinAsinCsinA

=sinAcosBsinAsinBcosAsinBsinAsinB+sinBcosCsinBsinCcosBsinCsinBsinC+sinCcosAsinCsinAcosCsinAsinCsinA

=cotBcotA+cosCcotB+cotAcotC

=0

= RHS

Hence proved.


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