A+B+C=π⇒A+B=π−C (1)B+C=π−A (2)A+C=π−B (3)
To Prove: sinA2+sinB2+sinC2−1=4sinπ−44.sinπ−B4.sinπ−C4LHS=sinA2+sinB2+sinC2−1=sinA2+sinB2+sinC2−sinπ2 (∵1=sinπ2)=2sinA2+B22cosA2−B22+2cosC2+π22sinC2−π22=2sinA+B4cosA−B4+2cosC+π4sinC−π4=2sinπ−C4cosA−B4+2cosC+π4(−sinπ−C4)=2sinπ−C4[cosA−B4−cosC+π4]=2sinπ−C4[−2sinA−B4+C+π42sinA−B4−C+π2]=−4sinπ−C4[sin(A+C)−B+π8sinA+π−(B+C)8]=−4sinπ−C4[sin(A+C)−B+π8sinA+π−(π−4)8]
=−4sinπ−C4[sin2π−2B8sin2A−2π8]=−4sinπ−C4[sinπ−B4sinA−π4]=4sinπ−C4 sinπ−B4 4sinπ−A4=RHS
Hence Proved.