Consider the surfaces: $$\tag{2} \begin{equation} \Sigma=\left\{\left(x, y, \log \left(\frac{\cos (x)}{\cos (y)}\right)\right) \in \mathbb{R}^3 \mid-\frac{\pi}{2}<x, y<+\frac{\pi}{2}\right\} \end{equation} $$ $$ \Sigma^2=\begin{equation} \left\{(x, y, z) \in \mathbb{R}^3 \mid x^2+y^2+z^2=1\right\} \end{equation} $$ I want to show that there does not exist a local isometry from $\Sigma$ to $\Sigma^2$. I believe we can use Gauss theorem, which says that the Gaussian curvature $K$ of a surface is invariant by local isometries. I found that the Gaussian curvature of $\Sigma$ is $$ \begin{equation} K_{\Sigma}=-\left(\frac{\sec x \sec y}{1+\tan ^2 x+\tan ^2 y}\right)^2 \end{equation}, $$ whereas the Gaussian curvature of the unit sphere, $\Sigma^2$ is everywhere equal to 1, $$ K_{\Sigma^2}=1 $$ Since there does not exist $(x,y) \in (-\pi/2,\pi/2)$ such that $K_\Sigma=K_{\Sigma^2}$, we can conclude from Gauss theorem that there does not exist a local isometry between the surfaces.
Is there anything wrong with my argument?