(b 2 -c 2 )/cos b+cos c+(c 2 -a 2 )/cos c+cos a+(a 2 -b 2 )/cos a+cos b=0

Question

Deepak Kumar Shringi · Answer

use cosine theroem replace all cos in form of sides a,b,c and you will get the answer

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(b 2 -c 2 )/cos b+cos c+(c 2 -a 2 )/cos c+cos a+(a 2 -b 2 )/cos a+cos b=0

(b²-c²)/cos b+cos c+(c²-a²)/cos c+cos a+(a²-b²)/cos a+cos b=0

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(b 2 -c 2 )/cos b+cos c+(c 2 -a 2 )/cos c+cos a+(a 2 -b 2 )/cos a+cos b=0

(b2-c2)/cos b+cos c+(c2-a2)/cos c+cos a+(a2-b2)/cos a+cos b=0

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(b²-c²)/cos b+cos c+(c²-a²)/cos c+cos a+(a²-b²)/cos a+cos b=0