a2 + b2 +c22abccos Acos BcosC

1.	a2 + b2 +c22abccos Acos BcosC
Answer» By cosine law of triangles,cos(A) = (b² + c² - a²)/2bccos(B) = (a² + c² - b²)/2accos(C) = (a² + b² - c²)/2ab iii) So, left side =(b² + c² - a²)/2abc + (a² + c² - b²)/2abc + (a² + b² - c²)/2abc = (b² + c² - a² + a² + c² - b² + a² + b² - c²)/2abc = (a² + b² + c²)/2abc = Right side [Proved]

Answer»

By cosine law of triangles,cos(A) = (b² + c² - a²)/2bccos(B) = (a² + c² - b²)/2accos(C) = (a² + b² - c²)/2ab

iii) So, left side =(b² + c² - a²)/2abc + (a² + c² - b²)/2abc + (a² + b² - c²)/2abc

= (b² + c² - a² + a² + c² - b² + a² + b² - c²)/2abc = (a² + b² + c²)/2abc = Right side [Proved]

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