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With the following method we can determinate the minimum and the maximum too for ALL REAL $m$ 1) If $m=1$, we have : $sinx*siny=cosx*cosy$, so $cos(x+y)=0,$thus $x+y=\pm\frac{\pi}{2}+2k\pi$, and in these situations we got: $E=2cos\left(\frac{\pi}{4}-x\right)$or $E=2sin\left(\frac{\pi}{4}-x\right)$. In both situation we have that $-\sqrt{2}\leq E\leq\sqrt{2}$ 2) If $m=-1$, we have: we work analogous, with the previouse demosntration. 3) If $\begin{vmatrix}m\end{vmatrix}>1$, we have : $\frac{a}{b}=\frac{c}{d}$$\Leftrightarrow\frac{a+b}{b-a}=\frac{c+d}{d-c}$, so from $\frac{sinx*siny}{cosx*cosy}=m,$we got $\frac{cos(x-y)}{cos(x+y)}=\frac{1+m}{1-m}$, so if we denote $(x-y)=2a,$ and $(x+y)=2b$, we have $\frac{cos(2a)}{1+m}=\frac{cos(2b)}{1-m}=k$ (notation), so $cos(2a)=k(1+m)$ and $cos(2b)=k(1-m)$ (1). But we have that $E^{2}=4cos^{2}acos^{2}b=(1+cos(2a))(1+cos(2b))=(1-m^{2})k^{2}+2k+1.$This quadratic function have maximum equal with $\frac{m^{2}}{m^{2}-1}$, so $E^{2}\leq$$\frac{m^{2}}{m^{2}-1}$, thus $-\frac{\begin{vmatrix}m\end{vmatrix}}{\sqrt{m^{2}-1}}\leq E\leq$ $\frac{\begin{vmatrix}m\end{vmatrix}}{\sqrt{m^{2}-1}}$. 4) If $\begin{vmatrix}m\end{vmatrix}<1$, we have: we can denote $m$=$\frac{1}{m'}$ and $\begin{vmatrix}m'\end{vmatrix}>1,$and we can proceeding analogous that in previouse demonstration, and we got $-\frac{\begin{vmatrix}m\end{vmatrix}}{\sqrt{1-m^{2}}}\leq E\leq$ $\frac{\begin{vmatrix}m\end{vmatrix}}{\sqrt{1-m^{2}}}$