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In a $\triangle ABC$, if $\tan A$, $\tan B$, $\tan C$ are in harmonic progression, then what is the minimum value of $\cot(B/2)$?

$\bf{My\; Try::}$ Here $A+B+C=\pi\;,$ Then $\tan A+\tan B+\tan C=\tan A\cdot \tan B\cdot \tan C$

and given $\tan A,\tan B\;,\tan C$ are in harmonic progression,

So we get $$\frac{2}{\tan B} = \frac{1}{\tan A}+\frac{1}{\tan C} = \frac{\tan A+\tan C}{\tan A\tan C}=\frac{\tan A\tan B\tan C-\tan B}{\tan A\tan C}$$

So $$\tan B -\frac{\tan B}{\tan A\tan C} = \frac{2}{\tan B}\Rightarrow \tan^2 B = \frac{2\tan A\tan C}{\tan A\tan C-1}$$

Now how can i solve after that, Help required, Thanks

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juantheron
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1 Answers1

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We have $\cot A+\cot C=2\cot B\ \ \ \ (1)$

and Prove that $\tan A + \tan B + \tan C = \tan A\tan B\tan C,$ $A+B+C = 180^\circ$ says $$\cot A\cot B+\cot B\cot C+\cot C\cot A=1$$

Eliminating $\cot C,$

we get $$\cot^2A-2\cot A\cot B+1-2\cot^2B=0$$ which a Quadratic Equation in $\cot A$

As $\cot A$ is real, the discriminant $$(2\cot B)^2-4(1-2\cot^2B)=4(3\cot ^2B-1)$$ must be $\ge0$

$$\implies\cot^2 B\ge\dfrac13\iff\tan^2B\le3\ \ \ \ (2)$$

If $\cot B<0,$ by $(1),$ at least one of $\cot A,\cot C$ must be $<0$

If $\cot A,\cot B<0, A.B>90^\circ\implies A+B>180^\circ$ which is impossible

$\implies B<90^\circ\implies\tan B\ge 0$

By $(2),\tan B\le\sqrt3\implies B\le60^\circ$

Can you take it from here?

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lab bhattacharjee
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