Math, asked by rk09051995p0owz3, 1 year ago

If A+B+C=180 prove that cos^2a+cos^2B+cos^2C=1-COSACOSBCOSC

Answers

Answered by TheLifeRacer
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

Answered by Vaishnavi20kulkarni
35

Answer:

Step-by-step explanation:

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