In a ABC- prove that cos Aa - cos Bb - cos Cc - a2 - b2 - c22abc

Using cosine rule, we have cosA=b^2+c^2 a^22bc #160; #160; cosB=c^2+b^2 a^22ac and cosC=a^2+b^2 c^22ab L.H.S =cosAa+cosBb+cosCc =(b ...

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Standard XII

Mathematics

General Solution of Trigonometric Equation

In a ABC, p...

Question

In a $△ A B C$ , prove that $\frac{c o s A}{a} + \frac{c o s B}{b} + \frac{c o s C}{c} = \frac{a^{2} + b^{2} + c^{2}}{2 a b c}$

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Solution

Using cosine rule, we have
$cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$ $cos B = \frac{c^{2} + b^{2} - a^{2}}{2 a c}$ and $cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$
L.H.S $= \frac{cos A}{a} + \frac{cos B}{b} + \frac{cos C}{c}$
$= \frac{(\frac{b^{2} + c^{2} - a^{2}}{2 b c})}{a} + \frac{(\frac{c^{2} + b^{2} - a^{2}}{2 a c})}{b} + \frac{(\frac{a^{2} + b^{2} - c^{2}}{2 a b})}{c}$
$= \frac{b^{2} + c^{2} - a^{2}}{2 a b c} + \frac{c^{2} + b^{2} - a^{2}}{2 a b c} + \frac{a^{2} + b^{2} - c^{2}}{2 a b c}$
$= \frac{b^{2} + c^{2} - a^{2} + a^{2} + b^{2} - c^{2} + a^{2} + b^{2} - c^{2}}{2 a b c}$
$= \frac{a^{2} + b^{2} + c^{2}}{2 a b c} =$ R.H.S

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In a △ABC, prove that cosAa+cosBb+cosCc=a2+b2+c22abc

Using cosine rule, we havecosA=b2+c2−a22bc cosB=c2+b2−a22ac and cosC=a2+b2−c22abL.H.S=cosAa+cosBb+cosCc=(b2+c2−a22bc)a+(c2+b2−a22ac)b+(a2+b2−c22ab)c=b2+c2−a22abc+c2+b2−a22abc+a2+b2−c22abc=b2+c2−a2+a2+b2−c2+a2+b2−c22abc=a2+b2+c22abc=R.H.S

In a $△ A B C$ , prove that $\frac{c o s A}{a} + \frac{c o s B}{b} + \frac{c o s C}{c} = \frac{a^{2} + b^{2} + c^{2}}{2 a b c}$