Question

cotA/2 cotB/2 cotC/2 are in A.p then prove that a,b,c are in A.p​

Answer 1

Step-by-step explanation:

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Answer 2

$a^2, b^2\ and\ c^2$ are in Arithmetic Progression

Step-by-step explanation:

Let us Consider $a^2, b^2\ and\ c^2$ in Arithmetic Progression just as CotA, CotB ,CotC are in Arithmetic Progression.

We know that Cot ($\frac {A}{2}$), Cot $(\frac {B}{2})$ & Cot $(\frac {C}{2})$ are in Arithmetic Progression.

So,

$2Cot (\frac {B}{2}) = Cot (\frac {A}{2}) + Cot (\frac {C}{2})$

Applying Trigonometric Ratios we get-

$2\frac {Cos (\frac {B}{2})}{Sin(\frac {B}{2})} = \frac {Cos (\frac {A}{2})}{Sin(\frac {A}{2})} + \frac {Cos (\frac {C}{2})}{Sin(\frac {C}{2})}$ """(1)

As we know that,  

$\frac {SinA}{a} = \frac {SinB}{b} = \frac {SinC}{c} = k$ [Constant Suppose]

Thus,

SinA = ka, SinB = kb, SinC = kc

and $CosA = \frac {b^2+c^2-a^2}{2bc}$, $CosB = \frac {a^2+c^2-b^2}{2ac}$ & $CosC = \frac {b^2+a^2+c^2}{2ab}$

Putting values in equation (1) we get-

$\frac {2}{kb} \times \frac {a^2+c^2-b^2}{2ac} = \frac {1}{ka} \times \frac {b^2+c^2-a^2}{2bc} = \frac {1}{kc} \times \frac {b^2+a^2+c^2}{2ab}$

⇒ $2a^2 + 2c^2 - 2b^2 = b^2 + c^2 - a^2 + b^2 + a^2 - c^2$

⇒ $2b^2 = a^2 + c^2$

Hence, $a^2, b^2\ and\ c^2$are in Arithmetic Progression.