Question

5. Prove that a/bc + cosA/a = b/ca

+cos B / b = c/ab +Cos C /c​

Answer 1

Given:

We are asked to prove that

$\frac{a}{bc}+\frac{cosA}{a}=\frac{b}{ca}+\frac{cosB}{b}=\frac{c}{ab} +\frac{cosC}{c}$

Solution:

Before solving let us recall the formulae,

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

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

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

(i)  $\frac{a}{bc}+\frac{cosA}{a}=\frac{a}{bc}+\frac{b^{2} +c^{2} -a^{2} }{2abc}$  =  $\frac{2a^{2} +b^{2} +c^{2} -a^{2} }{2abc}$  =  $\frac{a^{2} +b^{2} +c^{2}}{2abc}$     _ _ _ (1)

Now,  

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

$\frac{a^{2} +b^{2} +c^{2}}{2abc}$   =   $\frac{cosB}{b}+\frac{b}{ca}$  _ _ _(2)

Similarly,

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

$\frac{a^{2} +b^{2} +c^{2}}{2abc}$   =     c_ _ _(3)

Now from equations  (1) , (2) and (3) we can write

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

∴   $\frac{cosA}{a}+\frac{a}{bc}$    =   $\frac{cosB}{b}+\frac{b}{ca}$  =   $\frac{cosB}{b}+\frac{b}{ca}$

Hence Proved.

Answer 2

Answer:

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