Question

Let $A,B$ and $C$ represents the angles of $△ABC$. If $\tan \frac{A}{2},\tan \frac{B}{2},\tan \frac{C}{2}$ are in H.P., then the minimum value of $\cot \frac{B}{2}$ is

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\sqrt{3}$

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

Given : $\tan \frac{A}{2},\tan \frac{B}{2},\tan \frac{C}{2}$ are in H.P., so
$\cot \frac{A}{2},\cot \frac{B}{2},\cot \frac{C}{2}$ are in A.P.
Therefore,
$\cot \frac{A}{2}+\cot \frac{C}{2}=2\cot \frac{B}{2}$
From equation $\left(1\right)$, we get
$\Rightarrow \cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}=3\cot \frac{B}{2}\Rightarrow \cot \frac{A}{2}\cot \frac{C}{2}=3\cdots \left(2\right)$

Now, using A.M., G.M. on $\cot \frac{A}{2},\cot \frac{C}{2}$, we get
$\cot \frac{A}{2}+\cot \frac{C}{2}\ge 2\sqrt{\cot \frac{A}{2}\cot \frac{C}{2}}\Rightarrow 2\cot \frac{B}{2}\ge 2\sqrt{3}\therefore \cot \frac{B}{2}\ge \sqrt{3}$