Question

In $△ABC,\sum \left(\frac{{\sin }^{2}A+\sin A+1}{\sin A}\right)$ is always greater than

• A. $9$
• B. $3$
• C. $27$
• D. None of these

Answer 1

The correct option is A $9$

$\frac{{\sin }^{2}A+\sin A+1}{\sin A}=\sin A+1+\frac{1}{\sin A}$

$\Rightarrow (\sin A+\frac{1}{\sin A})+1\ge 2+1=3[\because x+\frac{1}{x}\ge 2]$

$\therefore \sum \left(\frac{{\sin }^{2}A+\sin A+1}{\sin A}\right)\ge 3+3+3=9$