If ABC is a triangle and tanA2,tanB2,tanC2 are in H.P then the minimum value of cotB2 is equal to

In a triangle A+B+C= π⇒cotA2.cotB2.cotC2=cotA2+cotB2+cotC2....(i) ⇒cotA2.cotB2.cotC2 are in AP⇒cotA2+cotc2=2cotB2 cotA2cotC2=3⇒GM of cot A2 and cot C2=cotA2cotC2=3 and AM of cot A2 and cot C2=cotA2+cotC22=cotB2. But AM≥GM⇒cotB2≥3

If ABC is a triangle and tanA2,tanB2,tanC2 are in H.P then the minimum value of cotB2 is equal to

In a triangle A+B+C= π⇒cotA2.cotB2.cotC2=cotA2+cotB2+cotC2....(i) ⇒cotA2.cotB2.cotC2 are in AP⇒cotA2+cotc2=2cotB2 cotA2cotC2=3⇒GM of cot A2 and cot C2=cotA2cotC2=3 and AM of cot A2 and cot C2=cotA2+cotC22=cotB2. But AM≥GM⇒cotB2≥3

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If ABC is a triangle and $\tan \frac{A}{2}, \tan \frac{B}{2}, \tan \frac{C}{2}$ are in H.P then the minimum value of $\cot \frac{B}{2}$ is equal to

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$- \sqrt{3}$

$\sqrt{3}$

$2$

$- 2$

answer is B.

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Detailed Solution

In a triangle A+B+C= $π$
$\Rightarrow \cot \frac{A}{2} . \cot \frac{B}{2} . \cot \frac{C}{2} = \cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} .... (i)$
$\Rightarrow \cot \frac{A}{2} . \cot \frac{B}{2} . \cot \frac{C}{2}$ are in AP
$\Rightarrow \cot \frac{A}{2} + \cot \frac{c}{2} = 2 \cot \frac{B}{2}$
$\cot \frac{A}{2} \cot \frac{C}{2} = 3 \Rightarrow G M$ of cot $\frac{A}{2}$ and cot $\frac{C}{2}$
$= \sqrt{\cot \frac{A}{2} \cot \frac{C}{2}} = \sqrt{3}$ and AM of cot $\frac{A}{2}$ and cot $\frac{C}{2}$
$= \frac{\cot \frac{A}{2} + \cot \frac{C}{2}}{2} = \cot \frac{B}{2} .$
But $A M \geq G M \Rightarrow \cot \frac{B}{2} \geq \sqrt{3}$