Question

(cos A)/a + (Cos B )/b+ (cos C)/c = (a 2 +b 2 + c 2 )/2abc

Answer 1

cosA/a=b2+c2-a2/2abc
cosB=c2+a2-c2/2abc
cosC=a2+b2-c3/2abc

adding above three we get
(cos A)/a + (Cos B )/b+ (cos C)/c = (a 2 +b 2 + c 2 )/2abc

Answer 2

Answer:

$\frac{cosA}{a}+\frac{cosB}{b}+\frac{cosC}{c}$ =$\frac{a^{2}+b^{2}+c^{2}}{2abc}$

Step-by-step explanation:

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By Cosine Rule :

i) $CosA =\frac{b^{2}+c^{2}-a^{2}}{2bc}$

ii)$CosB =\frac{a^{2}+c^{2}-b^{2}}{2ac}$

iii)$CosC =\frac{b^{2}+a^{2}-c^{2}}{2ac}$

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Now ,

$LHS = \frac{cosA}{a}+\frac{cosB}{b}+\frac{cosC}{c}$

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

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

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

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

=$RHS$

Therefore,

$\frac{cosA}{a}+\frac{cosB}{b}+\frac{cosC}{c}$ =$\frac{a^{2}+b^{2}+c^{2}}{2abc}$

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