Question

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

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

Answer 1

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

As $\tan \frac{A}{2},\tan \frac{B}{2},\tan \frac{C}{2}\to H.P$
$\therefore \tan \frac{B}{2}=\frac{2\tan \frac{A}{2}\tan \frac{C}{2}}{\tan \frac{A}{2}+\tan \frac{C}{2}}$
$=\frac{\frac{2\sin \frac{A}{2}}{\cos \frac{A}{2}}.\frac{\sin \frac{C}{2}}{\cos \frac{C}{2}}}{\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}+\frac{\sin \frac{C}{2}}{\cos \frac{C}{2}}}=\frac{2\sin \frac{A}{2}\sin \frac{C}{2}}{\sin (\frac{A}{2}+\frac{C}{2})}$
$=\frac{\cos (\frac{A}{2}-\frac{C}{2})-\cos (\frac{A}{2}+\frac{C}{2})}{\sin (\frac{A}{2}+\frac{C}{2})}$
$=\frac{\cos (\frac{A}{2}+\frac{C}{2})-\cos (\frac{\pi }{2}+\frac{B}{2})}{\sin (\frac{\pi }{2}-\frac{B}{2})}$
$=\frac{\cos (\frac{A}{2}-\frac{C}{2})-\sin \frac{B}{2}}{\cos \frac{B}{2}}$
$\therefore \sin \frac{B}{2}=\cos (\frac{A}{2}-\frac{C}{2})-\sin \frac{B}{2}$
$\Rightarrow 2\sin \frac{B}{2}=\cos (\frac{A}{2}-\frac{C}{2})$
$\Rightarrow \cot \frac{B}{2}<\sqrt{3}$