<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] 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
$$
f(\arcsin\sqrt{2/3})=4/(2/3)+9\cdot\frac{2}{3}=12
$$
</s>