Find the minimum of $\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}$ Let $0<x,y<\frac {\pi}{2}$ such that $\sin (x+y)=\frac 23$, then find the minimum of
$$\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}$$
A) $\frac 23$
B) $\frac 43$
C) $\frac 89$
D) $\frac {16}{9}$
E) $\frac{32}{27}$

My attempts:
I think that the all possible answers are wrong. Because, by Am-Gm inequality we have
$$\frac{\sin x}{\cos y}+\frac{\cos y}{\sin x}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}≥2+2=4.$$
But, Wolfram Alpha gives us a different result : The Global Minimum doesn't exist. However, ​the local minimum must be $6.$
But, still the problem is not solved. Because Wolfram's graph shows that the minimum can be less than $6$.
I also tried
Let $$\sin x=a,\cos y=b,\cos x=c,\sin y=d$$ with
$$ab+cd=\frac 23≥2\sqrt{abcd}\implies abcd≤\frac 19\\ a^2+c^2=b^2+d^2=1 $$
then I need
$$\min \left(\frac ab+\frac ba+\frac cd+\frac dc\right)$$
But, I can't do anything from here.  Finally, I attach the graph drawn by WA.


You are correct.  Let $$f(x,y) = g + \frac{1}{g} + h + \frac{1}{h}$$ where $$g(x,y) = \frac{\sin x}{\cos y}, \quad h(x,y) = \frac{\cos x}{\sin y}.$$  Then since $0 \le x, y \le \pi/2$, we must have $0 \le \sin x, \cos x, \sin y, \cos y \ge 1$, hence $g, h \ge 0$.  Then AM-GM proves $g + 1/g \ge 2$, as well as $h + 1/h \ge 2$, thus $f \ge 4$, and this is without the added constraint $\sin (x+y) = \frac{2}{3}$.  None of the answer choices is greater than $4$, so all are incorrect.
But also be advised that Wolfram Alpha is also misleading, because we have established that it is impossible for $f$ to be negative.  Thus there must be a lower bound.
The actual minimum is $6$.  To see this, let $c_1 = \arcsin \frac{2}{3} \approx 0.729728$, and $c_2 = \pi - c_1 \approx 2.41186$.  These comprise the two possible values for $x + y$ in the square $[0,\pi/2]^2$.  Then the transformation $u = x+y$, $v = x-y$ gives
$$f = 8\frac{\sin u \cos^2 v}{\cos 2v - \cos 2u} = 4 \frac{\sin u \cos^2 v}{ \cos^2 v + \sin^2 u - 1}$$  Then since $\sin u = 2/3$,
$$f = \frac{24\cos^2 v}{-5 + 9 \cos^2 v} = \frac{8}{3} + \frac{\frac{40}{27}}{t - \frac{5}{9}}$$ where $t = \cos^2 v$.  Consider the range of $t$.  Since $x + y \in \{c_1, c_2\}$ and $0 \le x, y \le \pi/2$, it follows that $0 \le |x-y| \le c_1$, and in turn, $$\frac{5}{9} \le t \le 1.$$  As $f$ is obviously a decreasing function of $t$ on this interval, the minimum is attained when $t$ is as large as possible, namely $t = 1$, corresponding to $v = 0$, or $x = y \in \{c_1/2, c_2/2\}$, and $$f = \frac{8}{3} + \frac{40/27}{4/9} = 6.$$

Here is a plot of $f(x,y)$ and the two curves for which $\sin (x+y) = \frac{2}{3}$.  As you can see, they do have a global minimum, and as $f$ is never negative, neither are the curves.

