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in a triangle of ABC, cos^2B+cos^2C-sin^2A-2cosAcosBcosC=0

in a triangle of ABC, cos^2B+cos^2C-sin^2A-2cosAcosBcosC=0

Grade:11

2 Answers

Ajay
209 Points
5 years ago
The question does not seem to be correct .
consisder an equilateral traingle wil all angles 60. 
Subtituting Cos 60 = ½  and Sin 60 = sqrt3/2  the LHS becomes
¼ + ¼ -3/4 – 2*1/2*1/2*1/2 which is not zero hence it cant be proved.
Ajay
209 Points
5 years ago
The correct question should be
cos^2B+cos^2C-sin^2A + 2cosAcosBcosC=0 for which proof can be given, but before that please confirm that there is mistake in question

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