Question

In any $\Delta ABC$, $\sum \frac{{\sin }^{2}A+\sin A+1}{\sin A}$ is always greater than:

• A. 0
• B. 6
• C. 8
• D. 7

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A 0
B 6
C 8
D 7

In $△ABC,\sum \frac{{\sin }^{2}A+\sin A+1}{\sin A}$

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

$=\sin A+\frac{1}{\sin A}+1+\sin B+\frac{1}{\sin B}+1+\sin C+\frac{1}{\sin C}+1$

$=\sin A+\frac{1}{\sin A}+\sin B+\frac{1}{\sin B}+\sin C+\frac{1}{\sin C}+3$

As $\sin \theta$ in any triangle is always greater than $0$

$\therefore \sin \theta +\frac{1}{\sin \theta }\ge 2$

$\therefore \sin A+\frac{1}{\sin A}+\sin B+\frac{1}{\sin B}+\sin C+\frac{1}{\sin C}\ge 6$

$\Rightarrow \sin A+\frac{1}{\sin A}+\sin B+\frac{1}{\sin B}+\sin C+\frac{1}{\sin C}+3\ge 6+3$

Thus,$\sum \frac{{\sin }^{2}A+\sin A+1}{\sin A}\ge 9$