In any ABC, if a^2, b^2, c^2 are in A.P., prove that , and are also in A.P.

a^2, b^2, c^2 are in A.P. -2a^2,-2b^2,-2c^2, are in A.P. (a^2+b^2+c^2)-2a,(a^2+b^2+c^2)-2b^2, a^2+b^2+c^2)-2c^2 are in A.P. (b^2+c^2-a^2),(c^2+a^2-b^2),(b^2+a^2-c^2) are in A.P. (b^2+c^2-a^2) 2abc , (c^2+a^2-b^2) 2abc , (b^2+a^2-c^2) 2abc are in A.P. 1 a (b^2+c^2-a^2) 2bc ,1 b (c^2+a^2-b^2) 2ac ,1 c (b^2+a^2-c^2) 2ab are in A.P. 1 a ,1 b ,1 c are in A.P. k a , k b , k c are in A.P. , , are in A.P. , , are in A.P.

In any ABC, if a^2, b^2, c^2 are in A.P., prove that , and are al…

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Question

In any $\triangle \text{ABC},$ if $a^2, b^2, c^2$ are in A.P., prove that $\cot\text{A},\cot\text{B}$ and $\cot\text{C}$ are also in A.P.

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Answer

$a^2, b^2, c^2$ are in A.P.$\Rightarrow-2\text{a}^2,-2\text{b}^2,-2\text{c}^2,$ are in A.P.
$\Rightarrow(\text{a}^2+\text{b}^2+\text{c}^2)-2\text{a},(\text{a}^2+\text{b}^2+\text{c}^2)-2\text{b}^2,\$\text{a}^2+\text{b}^2+\text{c}^2)-2\text{c}^2$ are in A.P.
$\Rightarrow(\text{b}^2+\text{c}^2-\text{a}^2),(\text{c}^2+\text{a}^2-\text{b}^2),(\text{b}^2+\text{a}^2-\text{c}^2)$ are in A.P.
$\Rightarrow\frac{(\text{b}^2+\text{c}^2-\text{a}^2)}{2\text{abc}},\frac{(\text{c}^2+\text{a}^2-\text{b}^2)}{2\text{abc}},\frac{(\text{b}^2+\text{a}^2-\text{c}^2)}{2\text{abc}}$ are in A.P.
$\Rightarrow\frac{1}{\text{a}}\frac{(\text{b}^2+\text{c}^2-\text{a}^2)}{2\text{bc}},\frac{1}{\text{b}}\frac{(\text{c}^2+\text{a}^2-\text{b}^2)}{2\text{ac}},\frac{1}{\text{c}}\frac{(\text{b}^2+\text{a}^2-\text{c}^2)}{2\text{ab}}$ are in A.P.
$\Rightarrow\frac{1}{\text{a}}\cos\text{A},\frac{1}{\text{b}}\cos\text{B},\frac{1}{\text{c}}\cos\text{C}$ are in A.P.
$\Rightarrow\frac{\text{k}}{\text{a}}\cos\text{A},\frac{\text{k}}{\text{b}}\cos\text{B},\frac{\text{k}}{\text{c}}\cos\text{C}$ are in A.P.
$\Rightarrow\frac{\cos\text{A}}{\sin\text{A}},\frac{\cos\text{B}}{\sin\text{B}},\frac{\cos\text{C}}{\sin\text{C}}$ are in A.P.
$\Rightarrow\cot\text{A},\cot\text{B},\cot\text{C}$ are in A.P.

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