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Question

ΔABC with sides a,b,c opposite to the angles A,B,C satisfies the equation
log2(2cosC2cosAB2)log2(sin(A+B))=1, then the value of (a2+b24c2+2ab) is

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Solution

log2(2cosC2cosAB2)log2(sin(A+B))=1
log22+log2(cosC2cosAB2)log2(sin(A+B))=1
log2(cosπ(A+B)2cosAB2)=log2(sin(A+B))
sinA+B2cosAB2=sin(A+B)
sinA+sinB2=sin(πC)
sinA+sinB=2sinC
​​​​​​​a2R+b2R=2c2R
​​​​​​​a+b=2c
a2+b24c2+2ab=(a+b)2(2c)2=0
​​​​​​​

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