Question

The minimum value of $\frac{1}{8}(16{\csc }^{2}x+9{\sin }^{2}x)$ is...........

Answer 1

$\frac{1}{8}(16{\csc }^{2}x+9{\sin }^{2}x)=2{\csc }^{2}x+\frac{9}{8}{\sin }^{2}x$
We know, $AM\ge GM\Rightarrow \frac{2{\csc }^{2}x+\frac{9}{8}{\sin }^{2}x}{2}\ge \sqrt{2{\csc }^{2}x\times \frac{9}{8}{\sin }^{2}x}\Rightarrow \frac{2{\csc }^{2}x+\frac{9}{8}{\sin }^{2}x}{2}\ge \sqrt{\frac{9}{4}}\Rightarrow \frac{2cs{c}^{2}x+\frac{9}{8}{\sin }^{2}x}{2}\ge \frac{3}{2}\Rightarrow 2cs{c}^{2}x+\frac{9}{8}{\sin }^{2}x\ge 3$
hence, its minimum value will be $3$