Question

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0​
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Answer 1

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof:

We know that, by Law of Cosines,

• $\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$

$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$

$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C +$

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

$\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$

$\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

On combining the fractions,

$\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}$

$\footnotesize \sf \dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Regrouping the terms,

$\sf \footnotesize\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}\\ \:$

$\sf \footnotesize\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}$

$\sf\longmapsto\dfrac{0}{2abc}$

$\mapsto\bf 0=RHS$

LHS = RHS proved.

Answer 2

Answer:

The given statement is proved.

Step-by-step explanation:

Let us take LHS and find the value of it.

$L.H.S.= \frac{b^2-c^2}{a}cosA+\frac{x^2-a^2}{b}cosB+\frac{a^2-b^2}{c}cosC$.

From the law of cosines, we use the formulae for $cosA, cosB$ and $cosC$.

Substituting the values in the equation, we get

$=\frac{b^2-c^2}{a}\frac{b^2+c^2-a^2}{2bc}+\frac{c^2-a^2}{b}\frac{c^2+a^2-b^2}{2ca}+\frac{a^2-b^2}{c}\frac{a^2+b^2-c^2}{2ab}$

Multiplying the values, we get

$=\frac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}+\frac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}+\frac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}$

$=\frac{(b^2-c^2)(b^2+c^2)-a^2(b^2-c^2)}{2abc}+\frac{(c^2-a^2)(c^2+a^2)-b^2(c^2-a^2)}{2abc}+\frac{(a^2-b^2)(a^2+b^2)-c^2(a^2-b^2)}{2abc}$

Now using the formula, $(x+y)(x-y)=x^2-y^2$

$=\frac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\frac{c^4-a^4-b^2c^2+a^2c^2}{2abc}+\frac{a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Now combining the fractions, since the denominators are same, we can add the numerators directly.

$=\frac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2c^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}$

Now we cancel all the similar terms with opposite symbols,

$=\frac{0}{2abc}$

Anything that divides zero, gives the value zero.

$=0$

$=R.H.S$

Therefore, the given statement is proved.