Question

In a triangle $ABC$, suppose that $\tan \frac{A}{2},\tan \frac{B}{2}$ and $\tan \frac{C}{2}$ are in harmonic progression.
What is the maximum possible value of angle $B$ ?

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $\frac{\pi }{3}$

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

But $\tan \frac{A}{2},\tan \frac{B}{2},\tan \frac{C}{2}$ are in H.P.
$\therefore \cot \frac{A}{2},\cot \frac{B}{2},\cot \frac{C}{2}$ are in A.P.
$\Rightarrow \cot \frac{A}{2}+\cot \frac{C}{2}=2\cot \frac{B}{2}$

Hence, Eq.$\left(1\right)$ becomes
$\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}=3\cot \frac{B}{2}$
$\Rightarrow \cot \frac{A}{2}\cot \frac{C}{2}=3$


$\frac{\cot \frac{A}{2}+\cot \frac{C}{2}}{2}\ge \sqrt{\cot \frac{A}{2}\cot \frac{C}{2}}$
$\Rightarrow \cot \frac{B}{2}\ge \sqrt{3}$
$\cot x$ is decreasing in $(0,\pi )$
$\Rightarrow \frac{B}{2}\le {\cot }^{-1}\left(\sqrt{3}\right)$
$\Rightarrow B\le \frac{\pi }{3}$
Thus, the maximum possible value of angle $B$ is $\frac{\pi }{3}$