Question

If ABC is a triangle and tanA/2, tanB/2, tanC/2 are in HP, then find the minimum value of cotB/2 is equal to?​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The minimum value of cot B/2 is equal to √3

Step-by-step explanation:

If ABC is triangle, then,

$A+B+C=\pi$

On dividing 2 on both sides, we get,

$\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2}$

$\Rightarrow \frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2}$

$\cot \left(\frac{A}{2}+\frac{B}{2}\right)=\cot \left(\frac{\pi}{2}-\frac{C}{2}\right)$

$\frac{\cot (A / 2) \cot (B / 2)-1}{\cot\left(\frac{A}{2}\right)+\cot \left(\frac{B}{2}\right)}=\tan \left(\frac{C}{2}\right)=\frac{1}{\cot \left(\frac{C}{2}\right)}$

$\cot \left(\frac{A}{2}\right) \cot \left(\frac{B}{2}\right) \cot \left(\frac{C}{2}\right)-\cot \left(\frac{C}{2}\right)=\cot \left(\frac{A}{2}\right)+\cot \left(\frac{B}{2}\right)$

$\cot \left(\frac{A}{2}\right) \cot \left(\frac{B}{2}\right) \cot \left(\frac{C}{2}\right)=\cot \left(\frac{C}{2}\right)+\cot \left(\frac{A}{2}\right)+\cot \left(\frac{B}{2}\right)$

Since, tan A/2, tan B/2, tan C/2 are in HP, then,

$\cot \left(\frac{A}{2}\right), \cot \left(\frac{B}{2}\right), \cot \left(\frac{C}{2}\right)$ will be in A.P

$\cot \left(\frac{A}{2}\right)+\cot \left(\frac{C}{2}\right)=2 \cot \left(\frac{B}{2}\right)$

Now, the above equation becomes,

$\cot \left(\frac{A}{2}\right) \cot \left(\frac{B}{2}\right) \cot \left(\frac{C}{2}\right)=2 \cot \left(\frac{B}{2}\right)+\cot \left(\frac{B}{2}\right)$

$\cot \left(\frac{A}{2}\right) \cot \left(\frac{C}{2}\right)=3$

G.M of cot A/2, cot C/2 is:

$\Rightarrow\sqrt{\cot \left(\frac{A}{2}\right)\times \cot \left(\frac{C}{2}\right)}=\sqrt{3}$

A.M of cot A/2, cot C/2 is:

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

GM ≥ AM

$\cot (B / 2)\geq\sqrt{3}$