Question

The minimum value of $\tan B\tan C$ in an acute angled triangle $ABC$ is

• A. $\tan \frac{A}{2}$
• B. $\cot \frac{A}{2}$
• C. $co{\sec }^2\frac{A}{2}$
• D. ${\cot }^2\frac{A}{2}$

Answer 1

The correct option is D ${\cot }^2\frac{A}{2}$

$\tan (B+C)=\frac{\tan B+\tan C}{1-\tan B\tan C}=-\tan A$
$\therefore \tan A=\frac{\tan B+\tan C}{\tan B\tan C-1}\ge \frac{2\sqrt{\tan B\tan C}}{\tan B\tan C-1}$
If $x=\tan B\tan C$
$x-1\ge 2\sqrt{x}\cot Aorx-2\sqrt{x}\cot A\ge 1$
Add ${\cot }^{2}A$ to both sides
${(\sqrt{x}-\cot A)}^2\ge co{\sec }^{2}A$
$\therefore \sqrt{x}\ge \cot A+co\sec A=\frac{1+\cos A}{\sin A}=\cot \frac{A}{2}\therefore x\ge {\cot }^2\frac{A}{2}$