Question

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0​

Answer 1

Prove that:

`\:\:\sf\:\:\left((b^2-c^2)/(a)\right)\cos A+\left((c^2-a^2)/(b)\right)\cos B +\left((a^2-b^2)/(c)\right)\cos C=0`

Proof:

We know that, by Law of Cosines,

• 
  `\sf \cos A=(b^2+c^2-a^2)/(2bc)`
• 
  `\sf \cos B=(c^2+a^2-b^2)/(2ca)`
• 
  `\sf \cos C=(a^2+b^2-c^2)/(2ab)`

Taking LHS

`\left((b^2-c^2)/(a)\right)\cos A+\left((c^2-a^2)/(b)\right)\cos B +\left((a^2-b^2)/(c)\right)\cos C`

Substituting the value of cos A, cos B and cos C,

`\longmapsto\left((b^2-c^2)/(a)\right)\left((b^2+c^2-a^2)/(2bc)\right)+\left((c^2-a^2)/(b)\right)\left((c^2+a^2-b^2)/(2ca)\right)+\left((a^2-b^2)/(c)\right)\left((a^2+b^2-c^2)/(2ab)\right)`

`\longmapsto\left(((b^2-c^2)(b^2+c^2-a^2))/(2abc)\right)+\left(((c^2-a^2)(c^2+a^2-b^2))/(2abc)\right)+\left(((a^2-b^2)(a^2+b^2-c^2))/(2abc)\right)`

`\longmapsto\left(((b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2))/(2abc)\right)+\left(((c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2))/(2abc)\right)+\left(((a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2))/(2abc)\right)`

`\longmapsto\left(((b^4-c^4)-(a^2b^2-a^2c^2))/(2abc)\right)+\left(((c^4-a^4)-(b^2c^2-a^2b^2))/(2abc)\right)+\left(((a^4-b^4)-(a^2c^2-b^2c^2))/(2abc)\right)`

`\longmapsto(b^4-c^4-a^2b^2+a^2c^2)/(2abc)+(c^4-a^4-b^2c^2+a^2b^2)/(2abc)+(a^4-b^4-a^2c^2+b^2c^2)/(2abc)`

On combining the fractions,

`\longmapsto((b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2))/(2abc)`

`\longmapsto(b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2)/(2abc)`

Regrouping the terms,

`\longmapsto((a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2))/(2abc)`

`\longmapsto((0)+(0)+(0)+(0)+(0)+(0))/(2abc)`

`\longmapsto(0)/(2abc)`

`\longmapsto\bf 0=RHS`

LHS = RHS proved.