Question

In $\Delta ABC$, the minimum value of $\frac{\sum {\cot }^2\frac{A}{2}\cdot {\cot }^2\frac{B}{2}}{\prod {\cot }^2\frac{A}{2}}$ is :

• A. 1
• B. 2
• C. 3
• D. Non existent

Answer 1

Answer: (A)

$\frac{\sum {\cot }^2\frac{A}{2}\cdot {\cot }^2\frac{B}{2}}{\pi {\cot }^2\frac{A}{2}}={\tan }^2\frac{A}{2}+{\tan }^2\frac{B}{2}+{\tan }^2\frac{C}{2}$
Now we have ${\left(\tan \frac{A}{2}-\tan \frac{B}{2}\right)}^2\ge 0$
${\tan }^2\frac{A}{2}+{\tan }^2\frac{B}{2}\ge 2{\tan }^2\frac{A}{2}\tan \frac{B}{2}$
Similarly write this inequality for other angles and odd all in equalities, we get $\sum \tan \frac{A}{2}\tan \frac{B}{2}=1$