<s>[INST] 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.
 [/INST] 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 
$$y_{\text{min}}=\frac{4}{\sin^2x}+9\sin^2 x=\frac{4}{2/3}+9(2/3)=\color{red}{12}$$
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