Finding the maximum and minimum values of $a^2\sin^2\theta + b^2\csc^2\theta$ 
How do I find the maximum and minimum values of the following?
$$a^2\sin^2\theta + b^2\csc^2\theta$$
Is the max value $\infty$?

I tried to find the minimum value by using A.M$\geq$G.M. inequality (is there any other way?) and got it as 2|a||b| which seem to work for all the positive values when a>b (I checked it on Desmos by putting different values of a and b) but when a$\le$b it just don't match in the graph. So, what am I doing wrong here? And what should be the correct answer?
 A: The function is periodic with period $\pi$ so let us consider $ 0\le x \le \pi.$
Case-1: When $a>b$ in  $$f(x)=a^2\sin^2 x+ b^2 \csc^2 x~~~(1)$$ then
$$f'(x)=2a^2 \sin x \cos x-2b^2 \cot x \csc^2 x.= 2 \cos x(a^2 \sin^4 -b^2)/\sin^2 x ~~~(2)$$
$f'(x)=0,$ gives $\cos x=0$ or $\sin^4 x = b^2/a^2 \Rightarrow \sin^2 x= b/a \Rightarrow \sin x = \pm \sqrt{b/a} \Rightarrow x_1 = \pi/2, ~ \mbox{or} ~ x_{2,3}=\pm \sin^{-1} \sqrt{b/a}$.
Next $$f''(x)= b^2(4 \cot^2x \csc^2 x + 2 \csc^2 x)+2a^2(\cos^2 x- \sin^2 x)~~~~~~(3).$$ Get $f''(x_1)=-2(a^2-b^2)< 0,$ so local max. at $x=\pi/2$. Use $\sin^2 x=b/a, \cos^2 x=(a-b)/a$ etc. in (3) to get $f''(x_{2,3})=8a(a-b)>0,$ so two local minima at $x=x_{2,3}.$ 
Hence when $a>b$, the local max. and min. are given as
$$f_{max}=f(\pi/2)=a^2+b^2, f_{min}f(x_{2,3})=2ab.$$
Case-2: When $a<b$, then $x_{2,3}$ do not exist and $f''(\pi/2)=-2(a^2-b^2)>0,$ so there exists only one local min. and $f_{min}=f(\pi/2)=a^2+b^2$
In both the cases the function $f(x)$ is positive and unbounded which can becomes $\infty$ at $x=0, \pi.$
A: Let the max/min values of $a^2\sin^2\theta + b^2\csc^2\theta$ be $k$. Thus:
$$a^2\sin^2\theta + b^2\csc^2\theta = k \implies a^2 \sin^4 \theta-k\sin^2 \theta+b^2=0$$
$$\Delta = 0: (-k)^2 - 4a^2b^2 = 0 \implies k^2 = 4a^2b^2, k = 2|a||b|$$
and there is no maximum value because the range of $\csc^2 \theta$ is $[1, \infty)$.
Restricting the domain to $[0, \pi)$, this means that there are two asymptotes at $x = 0, \pi$ and one local minimum at $x = \pi/2$.
