Question

Minimum value of $\frac{1}{2}(\cot \frac{A}{2}+3\tan \frac{A}{2})$ where $A\in (0,{180}^o)$ occurs when A equals

• A. ${60}^o$
• B. ${90}^o$
• C. ${120}^o$
• D. ${30}^o$

Answer 1

The correct option is A ${60}^o$

Let, $S=\frac{1}{2}(\cot \frac{A}{2}+3\tan \frac{A}{2})$
$\frac{dS}{dA}=\frac{1}{2}[\frac{-1}{2}{\csc }^2\frac{A}{2}+\frac{3}{2}{\sec }^2\frac{A}{2}]=0$
${\csc }^2\frac{A}{2}=3{\sec }^2\frac{A}{2}$
$\frac{1}{{\sin }^2\frac{A}{2}}=\frac{3}{{\cos }^2\frac{A}{2}}$
$\frac{{\sin }^2\frac{A}{2}}{{\cos }^2\frac{A}{2}}=\frac{1}{3}$
$\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}=\frac{1}{\sqrt{3}}$
$\tan \frac{A}{2}=\frac{1}{\sqrt{3}}$
$\therefore \frac{A}{2}=30°$
$\therefore A=60°$

Now,
$\frac{d^{2}S}{d{A}^2}=\frac{1}{2}[\csc \frac{A}{2}\csc \frac{A}{2}\cot \frac{A}{2}+3\sec \frac{A}{2}\sec \frac{A}{2}\tan \frac{A}{2}]$
${\left[\frac{d^{2}S}{d{A}^2}\right]}_{A=60}>0$
$\therefore$ At $A={60}^o$, S is minimum