Find the least value of $4\csc^{2} x+9\sin^{2} x$ 
Find the least value of $4\csc^{2} x+9\sin^{2} x$

$a.)\ 14 \ \ \ \ \ \ \ \ \ b.)\ 10 \\
c.)\ 11 \ \ \ \ \ \ \ \ \ \color{green}{d.)\ 12} $
$4\csc^{2} x+9\sin^{2} x \\
= \dfrac{4}{\sin^{2} x} +9\sin^{2} x \\
= \dfrac{4+9\sin^{4} x}{\sin^{2} x}  \\
= 13  \ \ \ \ \ \ \ \ \ \ \ \ (0\leq \sin^{2} x\leq 1)
$
But that is not in options.
I look for a short and simple way.
I have studied maths upto $12$th grade.
 A: $$\left(\frac{2}{\sin x}-3\sin x\right)^2\geq0$$
$$\iff\csc^2 x+9\sin^2 x\geq12$$ with equality achieved when $\sin^2x=2/3.$
A: You can use $\displaystyle4\csc^2 x+9\sin^2x=\big(\frac{2}{\sin x}-3\sin x\big)^2+12$
A: Let $y=4\csc^2x+9\sin^2 x$
$$y'=-8\csc^2x\cot x+9\sin2 x$$
$$y''=-8\csc^2x(-\csc^2 x)-8\cot x(-2\csc^2 x\cot x)+18\cos2 x$$
$$y''=8\csc^4x+16\csc^2\cot^2 x+18\cos2 x$$
for maxima or minima, $y'=0$ hence, $$-8\csc^2x\cot x+9\sin2 x=0$$ 
$$2\cos x\left(9\sin x-\frac{4}{\sin^3 x}\right)=0$$
$$\frac{\cos x}{\sin^3 x}\left(9\sin^4 x-4\right)=0$$
$\cos x=0\implies x=\pi/2$ 
or $9\sin^4x-4=0\iff \sin^2 x=\frac{2}{3}\ \ \ \ \ (\forall \ \ \ \sin x\ne 0)$
One, can easily check that minimum of $y$ occurs for  $\sin^2 x=\frac 23$ ($y''>0$), 
hence, substituting $\sin^2 x=\frac{2}{3}$ in $y$, the minimum value is 
$$y_{\text{min}}=\frac{4}{\sin^2x}+9\sin^2 x=\frac{4}{2/3}+9(2/3)=\color{red}{12}$$
A: By AM-GM inequality
$$4\csc^{2} x+9\sin^{2} x \geq 2 \sqrt{36\csc^{2} x\sin^{2} x}$$
A: Without nice tricks, you can use calculus. Consider
$$
f(x)=\frac{4}{\sin^2x}+9\sin^2x
$$
defined over $(0,\pi)$. Note that the general function is periodic of period $\pi$.
Then
$$
f'(x)=-\frac{8\cos x}{\sin^3x}+18\sin x\cos x
=2\frac{\cos x}{\sin^3x}(9\sin^4x-4)
=2\frac{\cos x}{\sin^3x}(\sqrt{3}\sin x-\sqrt{2})(\sqrt{3}\sin x+\sqrt{2})(3\sin^2x+2)
$$
In the given interval, $\sin x>0$; so the derivative vanishes for $x=\arcsin\sqrt{2/3}$, $x=\pi/2$ and $x=\pi-\arcsin\sqrt{2/3}$.
Since $\lim_{x\to0^+}f(x)=\lim_{x\to\pi^-}f(x)=\infty$, the function has a minimum at $\arcsin\sqrt{2/3}$ and $x=\pi-\arcsin\sqrt{2/3}$ (it has the same value at these points). Clearly
$$
f(\arcsin\sqrt{2/3})=4/(2/3)+9\cdot\frac{2}{3}=12
$$
