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Question

sinA+sinB+sinCsinA+sinBsinC=cotA2cotB2.

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Solution

sinA+sinB+sinCsinA+sinBsinC
=2sin(A+B2)cos(AB2)+2sin(A+B2cos(A+B2)2sin(A+B2)cos(AB2)2sin(A+B2cos(A+B2)
=cos(AB2)+cos(A+B2)cos(AB2)cos(A+B2)
=2cosA2cosB22sinA2sinB2
=cotA2cotB2


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