Determine the values of $x,y$ such that $\cos x \cos z \cos y +\sin x \sin y <0$ for all $z$ To solve a problem I need to determine the range of valus $x,y$ such that

$\cos x \cos z \cos y +\sin x \sin y <0 \ \forall z$.

I think it is sufficient to let $\cos z=1$ and then let $\cos z = -1$ and solve the system consists of these two inequalities. But I have got odd an odd answer after doing this. Is my method problematic? How to deal with this question?
 A: HINT
Note that
$$\cos x \cos z \cos y +\sin x \sin y <0\iff\cos x \cos y\left(\cos z  +\tan x \tan y \right)<0
$$
then consider $2$ cases


*

*$\cos x \cos y>0 \implies \cos z  +\tan x \tan y <0\implies \tan x \tan y <-1$

*$\cos x \cos y<0 \implies \cos z  +\tan x \tan y >0\implies \tan x \tan y >1$
and use product to sum identities.
A: You are on the right track.
After rendering $\cos z = \pm 1$, you have the following "bounds" which are to be turned into single trigonometric functions by the cosine identities for a sum and difference.  Giving only half the information here because I suspect you should know these:
$\cos x \cos y + \sin x \sin y = \cos (x-y)$
$-\cos x \cos y + \sin x \sin y = -\cos (???)$
Now break out your graph paper.  You know that cosines are zero when the argument is an odd multiple of $\pi/2$.  So graph a series of lines corresponding to these values of $x-y$ and $???$.  The plotted lines divide the plane into a grid of squares that are tilted relative to the axes.  Try out points in the interiors of some of these squares and note the pattern of squares within which both bounding cosines have the right sign.
A: Write
$$x=j\pi+\xi,\quad y=k\pi+\eta,\qquad |\xi|, \>|\eta|\leq{\pi\over2}\ .$$
Then the condition
$$\sin x\sin y<t\cos x\cos y\qquad(-1\leq t\leq1)$$
translates into
$$(-1)^{j+k}\sin\xi\sin\eta<(-1)^{j+k}t\cos\xi\cos\eta\qquad(-1\leq t\leq1)\ .$$
I shall do the case $j+k$ even and assume $|\xi|$, $|\eta|<{\pi\over2}$ for simplicity. Then we want
$$\tan\xi\tan\eta< t\qquad(-1\leq t\leq 1)\ ,$$
which is tantamount to $\tan\xi\tan\eta<-1$, so that $\xi$ and $\eta$ necessarily have different sign $\ne0$. If $\xi<0<\eta$ we obtain the condition  $-\tan|\xi|\tan\eta<-1$, or $$\tan|\xi|>\cot\eta=\tan\left({\pi\over2}-\eta\right)\ .$$
This implies $|\xi|>{\pi\over2}-\eta$, and leads to a triangular shape in the second quadrant. When $\eta<0<\xi$ we obtain this  shape reflected in the fourth quadrant.
The case $j+k$ odd has to be treated in the same way.
