Question

cos^2A+cos^2b+2cosA×cosB×cosc=sin^2c​

Answer 1

Answer: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 HELPS U

Answer 2

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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