If A+B+C=180 prove that cos^2a+cos^2B+cos^2C=1-COSACOSBCOSC
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Answered by
120
heya,
this is so easy .
A+B+C=180°
A+B=180°-C
cos(A+B)=Cos(180°-C)
CosA×CosB-SinA×SinB=-cosC
now,interchaging it's place
CosA×CosB+CosC=SinA×SinB
now squaring on both side .
then ,(CosA×CosB+CosC)^2=sin^2A×sin^2B
=Cos^2A×Cos^2B+2CosA×CosB×CosC=(1-Cos^A)*(1-Cos^2B)
=Cos^2A×Cos^2B+2CosA×CosB×CosC+cos^2C=1-Cos^2B+Cos^2A-Cos^2A×Cos^2B
=Cos^2A+cos^2B+Cos^2C=1-2cosAcosBcosC
=hope it help you.
@rajukumar
this is so easy .
A+B+C=180°
A+B=180°-C
cos(A+B)=Cos(180°-C)
CosA×CosB-SinA×SinB=-cosC
now,interchaging it's place
CosA×CosB+CosC=SinA×SinB
now squaring on both side .
then ,(CosA×CosB+CosC)^2=sin^2A×sin^2B
=Cos^2A×Cos^2B+2CosA×CosB×CosC=(1-Cos^A)*(1-Cos^2B)
=Cos^2A×Cos^2B+2CosA×CosB×CosC+cos^2C=1-Cos^2B+Cos^2A-Cos^2A×Cos^2B
=Cos^2A+cos^2B+Cos^2C=1-2cosAcosBcosC
=hope it help you.
@rajukumar
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35
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