Question

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

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

Answer 1

Answer: (C)

$27$

$\prod \left(\frac{{\sin }^{2}A+\sin A+1}{\sin A}\right)$
$\because \sin A+\frac{1}{\sin A}\ge 2$ if $\angle A={90}^0$
$\Rightarrow \sin B+\frac{1}{\sin B}>2$
and $\sin C+\frac{1}{\sin C}>2$
$\therefore \prod \left(\frac{{\sin }^{2}A+\sin A+1}{\sin A}\right)$
$=(\sin A+1+\frac{1}{\sin A})(\sin B+1+\frac{1}{\sin B})(\sin C+1+\frac{1}{\sin C})$
Re-arranged as
$=(\sin A+\frac{1}{\sin A}+1)(\sin B+\frac{1}{\sin B}+1)(\sin C+\frac{1}{\sin C}+1)$
$>(2+1)(2+1)(2+1)$
$>3\times 3\times 3=27$
$\therefore \prod \left(\frac{{\sin }^{2}A+\sin A+1}{\sin A}\right)>27$