If A+B+C = 2S = sin(S-A) + sin(S-B) + sin(S-C) - sins
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Answer:
here is ur ans.
Given,
A+B+C=2S...........equ.1
now,
L.H.S.=sin(S-A)sin(S-B) + sinSsin(S-C)
=1/2[2sin(S-A)sin(S-B) + 2sinSsin(S-C)]
=1/2[{cos(S-A+S-B) - cos(S-A-S+B)} +{cos(S+S-C) -cos(S-S+C)}]
=1/2[cos(2S-A-B) - cos(B-A) + cos(2S-C) - cosC]
=1/2[cosC - cosC +{cos(2S-C) - cos(B-A)}]
=1/2[cos(A+B) - cos(B-A)].......from equ.1 2S-C=A+B
=1/2[2{sin(A+B+B-A)/2}{sin(A+B-A+B)/2}]
=(sin2B/2)(sin2A/2)
=sinBsinA
=R.H.S.
Step-by-step explanation:
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