Let a = b = c =k ...(1) Consider the LHS of the equation a +b +c . a +b +c =k( + + ) = k 2 (2 +2 +2 ) = k 2 ( 2A+ 2B+ 2C) = k 2 [2 (A + B) (A-B)+2 ] = k 2 [2 ( - C) (A-B)+2 ] = k 2 [2 (A-B)+2 ] = 2k 2 [ (A-B)+ ] =k [ (A-B)+ -(A + B) ] =k [ (A-B)- (A + B)] =k [2 ] =2k ...(1) Now, on putting k = C in equation (1), we get: 2c and on putting k = b in equation (1), we get: 2b So, from (1), we have a + b + c =2b =2c . Hence proved.

a + b + c =2b =2c

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$\text{a}\cos\text{A + b}\cos\text{B + c}\cos\text{C}=2\text{b}\sin\text{A}\sin\text{C}=2\text{c}\sin\text{A}\sin\text{B}$

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Answer

Let $\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}=\text{k }...(1)$
Consider the LHS of the equation $\text{a}\cos\text{A}+\text{b}\cos\text{B}+\text{c}\cos\text{C}.$
$\text{a}\cos\text{A}+\text{b}\cos\text{B}+\text{c}\cos\text{C}$
$=\text{k}(\sin\text{A}\cos\text{A}+\sin\text{B}\cos\text{B}+\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}(2\sin\text{A}\cos\text{A}+2\sin\text{A}\cos\text{A}+2\sin\text{C}\cos\text{C})$
$=\frac{\text{k}}{2}(\sin2\text{A}+\sin2\text{B}+\sin2\text{C})$
$=\frac{\text{k}}{2}[2\sin(\text{A + B})\cos(\text{A}-\text{B})+2\sin\text{C}\cos\text{C}]$
$=\frac{\text{k}}{2}[2\sin(\pi - \text{C})\cos(\text{A}-\text{B})+2\sin\text{C}\cos\text{C}]$
$=\frac{\text{k}}{2}[2\sin\text{C}\cos(\text{A}-\text{B})+2\sin\text{C}\cos\text{C}]$
$=\frac{2\text{k}\sin\text{C}}{2}[\cos(\text{A}-\text{B})+\cos\text{C}]$
$=\text{k}\sin\text{C}[\cos(\text{A}-\text{B})+\cos\{\pi-(\text{A + B})\}]$
$=\text{k}\sin\text{C}[\cos(\text{A}-\text{B})-\cos(\text{A + B})]$
$=\text{k}\sin\text{C}[2\sin\text{A}\sin\text{B}]$
$=2\text{k}\sin\text{A}\sin\text{B}\sin\text{C }...(1)$
Now,
on putting $\text{k}\sin\text{C = C}$ in equation (1), we get:
$2\text{c}\sin\text{A}\sin\text{B}$
and on putting $\text{k}\sin\text{B = b}$ in equation (1), we get:
$2\text{b}\sin\text{A}\sin\text{C}$
So, from (1), we have
$\text{a}\cos\text{A + b}\cos\text{B + c}\cos\text{C}=2\text{b}\sin\text{A}\sin\text{C}=2\text{c}\sin\text{A}\sin\text{B}.$
Hence proved.

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