Question

In any triangle ABC, prove that
(b ^ 2 - c ^ 2)/(a ^ 2) * cosA + (c ^ 2 - a ^ 2)/(b ^ 2) * cosB + (a ^ 2 - b ^ 2)/(c ^ 2) * cosC = 0​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\bigg(\dfrac{b ^{2} - c ^{2}}{a}\bigg)\,cos(A)\,+\,\bigg(\dfrac{c^{2} -a^{2}}{b}\bigg) \,cos(B)\,+\,\bigg(\dfrac{a^{2} -b^{2}}{c}\bigg)\,cos(C)}$

$\tt{=\bigg(\dfrac{b ^{2} - c ^{2}}{a}\bigg)\cdot\bigg(\dfrac{b ^{2}+c ^{2}-a^{2}}{2bc}\bigg)\,+\,\bigg(\dfrac{c^{2} -a^{2}}{b}\bigg)\cdot\bigg(\dfrac{a^{2}+c ^{2}-b^{2}}{2ac}\bigg)\,+\,\bigg(\dfrac{a^{2} -b^{2}}{c}\bigg)\cdot\bigg(\dfrac{a^{2}+b ^{2}-c^{2}}{2ab}\bigg)}$$\tt{=\dfrac{1}{abc}\{(b ^{2} - c ^{2})(b ^{2}+c ^{2}-a^{2})\,+\,(c^{2} -a^{2})(a^{2}+c ^{2}-b^{2})\,+\,(a^{2} -b^{2})(a^{2}+b ^{2}-c^{2})\}}$

$\tt{=\dfrac{1}{abc}\{b ^{4} - c ^{4}-a^{2}b^{2}+c^{2}a^{2}\,+\,c^{4} -a^{4}-b^{2}c^{2}+a^{2}b^{2}+\,a^{4} -b^{4}-c^{2}a^{2}+c^{2}b^{2}\}}$

$\tt{=\dfrac{1}{abc}\{b ^{4} - c ^{4}+c^{4} -a^{4}+a^{4} -b^{4}-a^{2}b^{2}+a^{2}b^{2}+c^{2}a^{2}-c^{2}a^{2}-b^{2}c^{2}+b^{2}c^{2}\}}$

$=\sf{0}$