Sin^2A+Sin^2B+Sin^2C = 2+2CosACosBCosC when A+B+C=180°
Answers
Answered by
58
Given A+B+C = 180
C = 180 -( A+B)
cos C = cos 180 - (A +B) = - cos (A+B)
now, Sin^2A + 1 - Cos^2B + Sin^2C
1 -(Cos^2B - Sin^2A) + Sin ^2C
1 - Cos (A+B) Cos (A - B) + 1- Cos^2C
2 + CosC [Cos(A -B) + Cos ( A + B)]
2 + 2 Cos A CosB CosC
Hence L.H.S = R.H.S
may this will help us mark as brainliest
Answered by
4
Answer:
sin^2A+sin^2B+sin^2C-2cosAxosBcosC
= 1-cos^2A+1-cos^2B+1-cos^2C-2cosAcosBcosC
= 3-cos^2A-cos^2B-cos^2C-2cosAcosBcosc
=1/2(6-2cos^2A-2cos^2B-2cos^2C-4cosAcosB
cos C
=then do it yourself I am going to sleeping
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