Question

In a $△ABC$, $\sum \frac{{\sin }^{2}C+\sin C+1}{\sin C}$ is always greater than

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

Answer 1

The correct option is C $9$

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

$=\left(\sin C+\frac{1}{\sin C}\right)+1\ge 2+1=3$ $[\because A.M\ge G.M\Rightarrow x+\frac{1}{x}\ge 2]$

Therefore, in a triangle $\sum \frac{{\sin }^{2}C+\sin C+1}{\sin C}\ge 3+3+3=9$